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Full text of "Mathematical Analysis"

Mathematical 
Analysis 

a straightforward 
approach 









K.G. BIN MORE 



This is an unf ussy account of mathematical 
analysis for first year courses at universities and 
polytechnics that places little reliance on the 
reader's previous mathematical background. 

The concepts involved are profound and subtle 
and their application in particular instances 
requires a mastery of technique which comes 
only with time and experience. It is tempting to 
introduce the concepts at a relatively high level of 
abstraction; it then becomes possible to expose 
the elegance of the underlying theory. But such 
treatments make great demands on a student's 
technical facility. At the other extreme is the 
danger of overwhelming students with technical 
detail. 

Professor Binmore's approach is to confine 
attention to 'bread and butter' analysis on the real 
line and, within this framework, to discuss only 
those technicalities which are immediately 
relevant to the basic concepts, the object being to 
promote confidence and understanding rather 
than awe or mindless computational efficiency. 
Emphasis is placed on the importance of tackling 
problems and the exercises are intended as an 
integral part of the text. Their worked solutions, 
which may be consulted when difficulty arises in 
solving a problem, occupy approximately a third 
of the book. This abundance of exercises and 
solutions makes the text highly suitable for those 
who want to learn the subject on their own as 
well as for those attending formal classes. 



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II 



MATHEMATICAL 
ANALYSIS 

A STRAIGHTFORWARD APPROACH 



K.G.BINMORE 

Professor of Mathematics L.S.E. 



CAMBRIDGE UNIVERSITY PRESS 

CAMBRIDGE 

LONDON - NEW YORK - MELBOURNE 



Published by the Syndics of the Cambridge University Press 

The Pitt Building, Trumpington Street, Cambridge CB2 1 RP 

Bentley House, 200 Euston Road, London NW1 2DB 

32 East 57th Street, New York, NY 10022, USA 

296 Beaconsfield Parade, Middle Park, Melbourne 3206, Australia 

© Cambridge University Press 1977 

First published 1977 

Printed in Great Britain at the University Press, Cambridge 

Library of Congress Cataloguing in Publication Data 

Binmore,KG 1940- 
Mathemalical analysis. 

Includes index. 

1. Mathematical analysis. I. Title. 
QA300.B536 515 76-28006 
ISBN 521 21480 7 hard covers 
ISBN 521 29167 4 paperback 



CONTENTS 








Preface 


IX 


1 


Real numbers 


1 


1.1 


Set notation 


1 


1 


2 


The set of real numbers 


2 


1 


3 


Arithmetic 


3 


1 


4 


Inequalities 


3 


1 


9 


Roots 


6 


1 


10 


Quadratic equations 


6 


1 


13- 


Irrational numbers 


9 


1 


14 


Modulus 


10 


2 


Continuum Property 


12 


2.1 


Achilles and the tortoise 


12 


2.2 


The continuum property 


13 


2.6 


Supremum and infimum 


15 


2.7 


Maximum and minimum 


15 


2.9 


Intervals ■:•> 


16 


2.11 


Manipulations with sup and inf 


18 


3 


Natural numbers 


20 


3.1 


Introduction 


20 


3.2 


Archimedean property 


20 


3.7 


Principle of induction 


22 


4 


Convergent sequences 


26 


4.1 


The bulldozers and the bee 


26 


4.2 


Sequences 


27 


4.4 


Definition of convergence 


27 


4.7 


Criteria for convergence 


30 


4.15 


Monotone sequences 


33 


4.21 


Some simple properties of convergent sequences 


37 


4.26 


Divergent sequences 


38 


5 


Subsequences 


41 


5.1 


Subsequences 


41 


5.8 


Bolzano-Weierstrass theorem 


46 



VI 


Contents 


™ * 




Contents 


vii 


5.12 


Lim sup and lim inf 


47 


12 


Monotone functions 


108 


5.16 


Cauchy sequences 


49 


12.1 


Definitions 


108 








12.3 


Limits of monotone functions 


108 


6 


Series 


53 


12.6 


Differentiable monotone functions 


110 


6.1 


Definitions 


53 


12.9 


Inverse functions 


110 


6.4 


Series of positive terms 


54 


12.11 


Roots 


112 


6.7 


Elementary properties of series 


56 


12.13 


Convex functions 


114 


6.12 


Series and Cauchy sequences 


57 


13 


Integration 




6.20 


Absolute and conditional convergence 


60 


119 


6.23 


Manipulations with series 


61 


13.1 


Area 


119 








13.2 


The integral 


120 


7 


Functions 


64 


13.3 


Some properties of the integral 


121 


7.1 


Notation 


64 


13.9 


Differentiation and integration 


124 


7.6 


Polynomial and rational functions 


66 


13.16 


Riemann integral 


127 


7.9 


Combining functions 


68 


13.19 


More properties of the integral 


129 


7.11 


Inverse functions 


68 


13.27 


Improper integrals 


132 


7.13 


Bounded functions 


70 


13.31 


Euler-Maclaurin summation formula 


134 


8 


Limits of functions 


74 


14 


Exponential and logarithm 


137 








14.1 - 


Logarithm 


137 


8.1 


Limits from the left 


74 


14.4 


Exponential 


140 


8.2 


Limits from the right 


74 


14.6 


Powers 


141 


8.3 


/(x)^/as:c-*£ 


75 








8.6 


Continuity at a point 


77 


15 


Power series 


143 


8.8 


Connexion with convergent sequences 


78 


15.1 


Interval of convergence 


143 


8.11 


Properties of limits 


79 


15.4 


Taylor series 


145 


8.16 
8.18 


Limits of composite functions 
Divergence 


81 
82 


15.7 


Continuity and differentiation 


147 








16 


Trigonometric functions 


150 


9 


Continuity 


84 


16.1 


Introduction 


150 


9.1 


Continuity on an interval 


84 


16.2 


Sine and cosine 


151 


9.7 


Continuity property 


86 


16.4 


Periodicity 


153 


10 


Differentiation 


91 


17 


The gamma function 


156 


10.1 


Derivatives 


91 


17.1 


Stirling's formula 


156 


10.2 


Higher derivatives 


92 


17.3 


The gamma function 


157 


10.4 


More notation 


93 


17.5 


Properties of the gamma function 


158 


10.5 


Properties of differentiable functions 


95 


18 






10.12 


Composite functions 


98 


Appendix 


161 










This contains the proofs of 'propositions' left unproved in the main 


11 


Mean value theorems 


100 


. 


body of the text. 




11.1 


Local maxima and minima 


100 




Solutions to exercises 


172 


11.3 


Stationary points 


101 








11.5 


Mean value theorem 


102 




Suggested further reading 


251 


11.9 


Taylor's theorem 


105 




Notation 
Index 


253 

255 



PREFACE 



This book is intended as an easy and unfussy introduction to mathematical 
analysis. Little formal reliance is made on the reader's previous mathematical 
background, but those with no training at all in the elementary techniques of 
calculus would do better to turn to some other book. 

An effort has been made to lay bare the bones of the theory by eliminating as 
much unnecessary detail as is feasible. To achieve this end and to ensure that all 
results can be readily illustrated with concrete examples, the book deals only 
with 'bread and butter' analysis on the real line, the temptation to discuss gener- 
alisations in more abstract spaces having been reluctantly suppressed. However, 
the need to prepare the way for these generalisations has been kept well in mind. 

It is vital to adopt a systematic approach when studying mathematical analy- 
sis. In particular, one should always be aware at any stage of what may be 
assumed and what has to be proved. Otherwise confusion is inevitable. For this 
reason, the early chapters go rather slowly and contain a considerable amount of 
material with which many readers may already be familiar. To neglect these 
chapters would, however, be unwise. 

The exercises should be regarded as an integral part of the book. There is a 
great deal more to be learned from attempting the exercises than can be obtained 
from a passive reading of the text. This is particularly the case when, as may fre- 
quently happen, the attempt to solve a problem is unsuccessful and it is necess- 
ary to turn to the solutions provided at the end of the book. 

To help those with insufficient time at their disposal to attempt all the exer- 
cises, the less vital exercises have been marked with the symbol t- (The same 
notation has been used to mark one or two passages in the text which can be 
omitted without great loss at a first reading.) The symbol * has been used to 
mark exercises which are more demanding than most but which are well worth 
attempting. 

The final few chapters contain very little theory compared with the number 
of exercises set. These exercises are intended to illustrate the power of the tech- 
niques introduced earlier in the book and to provide the opportunity of some re- 
vision of these ideas. 

This book arises from a course of lectures in analysis which is given at the 
London School of Economics. The students who attend this course are mostly 
not specialist mathematicians and there is little uniformity in their previous 



IX 



Preface x 

mathematical training. They are, however, quite well-motivated. The course is a 
'one unit' course of approximately forty lectures supplemented by twenty in- 
formal problem classes. I have found it possible to cover the material of this 
book in some thirty lectures. Time is then left for some discussion of point set 
topology in simple spaces. The content of the book provides an ample source of 
examples for this purpose while the more general theorems serve as reinforce- 
ment for the theorems of the text. 

Other teachers may prefer to go through the material of the book at a more 
leisurely pace or else to move on to a different topic. An obvious candidate for 
further discussion is the algebraic foundation of the real number system and the 
proof of the Continuum Property. Other alternatives are partial differentiation, 
the complex number system or even Lebesgue measure on the line. 

I would like to express my gratitude to Elizabeth Boardman and Richard 
Holmes for reading the text for me so carefully. My thanks are also due to 
'Buffy' Fennelly for her patience and accuracy in preparing the typescript. Final- 
ly, I would like to mention M.C. Austin and H. Kestelman from whom I learned 
so much of what I know. 



July 1976 



K.G.B. 



1 



REAL NUMBERS 



1.1 Set notation 

A set is a collection of objects which are called its elements. If x is an 
element of the set S, we say that x belongs to S and write 

xes. 

Ify does not belong to 5, we write y £ S. 

The simplest way of specifying a set is by listing its elements. We use the 
notation 

A = &1, >£,«,*} 

to denote the set whose elements are the real numbers \, 1 , \Jl, e and it. 
Similarly 

B = {Romeo, Juliet} 

denotes the set whose elements are Romeo and Juliet. 

This notation is, of course, no use in specifying a set which has an infinite 
number of elements. Such sets may be specified by naming the property which 
distinguishes elements of the set from objects which are not in the set. For 
example, the notation 

C = {x : x > 0} 

(which should be read 'the set of all x such that x > 0') denotes the set of all 
positive real numbers. Similarly 

D = {y : y loves Romeo} 

denotes the set of all people who love Romeo. 

It is convenient to have a notation for the empty set 0. This is the set which 
has no elements. For example, if x denotes a variable which ranges over the set 
of all real numbers, then 

{x : x 2 + 1 = 0} = 0. 

This is because there are no real numbers x such that x 2 = — 1 . 
If 5 and T are two sets, we say that 5 is a subset of T and write 



Real numbers 



SCT 



if every element of S is also an element of T. 

As an example, consider the sets P = {1 , 2, 3 , 4} and Q = {2, 4}. Then QCP. 
Note that this is not the same thing as writing Q e P, which means that Q is an 
element of P. The elements ofP are simply 1,2,3 and 4. But Q is not one of 
these. 

The sets A , B, C and D given above also provide some examples. We have 
A C Cand (presumably) BCD. 



1 .2 The set of real numbers 

It will be adequate for these notes to think of the real numbers as being 
points along a straight line which extends indefinitely in both directions. The 
line may then be regarded as an ideal ruler with which we may measure the 
lengths of line segments in Euclidean geometry. 



n/2 



-2 



The set of all real numbers will be denoted by R . The table below distin- 
guishes three important subsets of U. 



Subset 



Notation 



Elements 



Natural numbers 
(or whole numbers) 

Integers 

Rational numbers 
(or fractions) 



Q 



1,2,3,4,5,. .. 

. .-2,-1,0, 1,2,3, 



0,1,2.-1.*, If. -I, -I 



Not all real numbers are rational. Some examples of irrational numbers are 
v2, e and ir. 

While we do not go back to first principles in these notes, the treatment will 
be rigorous in so far as it goes. It is therefore important to be clear, at every 
stage, about what our assumptions are. We shall then know what has to be 
proved and what may be taken for granted. Our most vital assumptions are con- 
cerned with the properties of the real number system. The rest of this chapter 
and the following two chapters are consequently devoted to a description of the 



Real numbers 

properties of the real number system which we propose to assume and to some 
of their immediate consequences. 



1.3 Arithmetic 

The first assumption is that the real numbers satisfy all the usual laws 
of addition, subtraction, multiplication and division. 

The rules of arithmetic, of course, include the proviso that division by zero is 
not allowed. Thus, for example, the expression 

2 


makes no sense at all. In particular, it is not true that 



We shall have a great deal of use for the symbol °°, but it must clearly be 
understood that °° does not represent a real number. Nor can it be treated as 
such except in very special circumstances. 



1 .4 Inequalities 

The next assumptions concern inequalities between real numbers and 
their manipulation. 

We assume that, given any two real numbers a and b, there are three mutually 
exclusive possibilities: 

(i) a > b (a is greater than b) 

(ii) a = b (a equals b) 

(iii) a < b (a is less than b). 

Observe that a < b means the same thing as b > a. We have, for example, the 
following inequalities. 

1 >0; 3>2; 2<3; -K0; -3<-2. 
There is often some confusion about the statements 

(iv) a~>b (a is greater than or equal to b) 

(v) a < b (a is less than or equal to b). 
To clear up this confusion, we note that the following are all true statements. 

1 >0\ 3>2; 1 > 1; 2<3; -1 <0; -3«— 3. 
We assume four basic rules for the manipulation of inequalities. From these 



4 



Real numbers 



the other rules may be deduced. 
(I) If a >b and b >c, then a >c. 



(II) If a > b and c is any real number, then 

a+c>b+c. 

(III) If a > b and c > 0, then ac > 6c (i.e. inequalities can be multiplied 
through by a positive factor). 

(IV) If a > b and c < 0, then ac < be (i.e. multiplication by a negative factor 
reverses the inequality). 



7.5 Example If a > 0, prove that a -1 > 0. 

/Voo/ We argue by contradiction. Suppose that a > but that a -1 < 0. 
It cannot be true that a" 1 = (since then = O.a = 1). Hence 

a _1 <0. 

By rule III we can multiply this inequality through by a (since a > 0). Hence 

1 = a"'.a<0.a = 0. 

But 1 < is a contradiction. Therefore the assumption a -1 < was false. Hence 
a _I >0. 



1.6 Example If x and y are positive, thenar <y if and only if x 2 <y 2 . 

Proof We have to show two things. First, that x < y implies x 2 < y 2 , 
and secondly, that x 2 < y 2 implies x < y. 

(i) We begin by assuming that* <y and try to deduce that x 2 <y 2 . Multiply 
the inequality x < y through by x > (rule III). We obtain 

x 2 <xy. 

Similarly 

xy<y 2 . 

But now* 2 <y 2 follows from rule I. 

(ii) We now assume that x 2 <y 2 and try and deduce that* < y. Adding —x 2 
to both sides of* 2 < y 2 (rule II), we obtain 

y 2 -x 2 >0 

i.e. (y~x)(y + x)>0. (1) 



Real numbers • 

Since x + y > 0, (x +y)~ 1 > (example 1 .6). We can therefore multiply 
through inequality (1) by (x + y)' 1 to obtain 

y-x>0 

i.e. x<y. 

(Alternatively, we could prove (ii) as follows. Assume that* 2 <y 2 but that 
x > y. From x > y it follows (as in (i)) that x 2 > y 2 , which is a contradiction.) 



1.7 Example Suppose that, for any e>0, a <b + e. Thena <fc. 

Proof Assume that a > b. Then a — b > 0. But, for any e > 0, 

a < b + e. Hence a < b + e in the particular case when e = b —a. Thus 

a<b + (a-b) 

and so a<a. 

This is a contradiction. Hence our assumption a > b must be false. Therefore 
a<b. 

(Note: The symbol e in this example is the Greek letter epsilon. It should be 
carefully distinguished from the 'belongs to' symbol G and also from the symbol 
% which is the Greek letter*/'.) 



1.8 Exercise 

(1) If* is any real number, prove that* 2 >Q. If 0<a< 1 andb > 1, 
prove that 

(i)0<a 2 <a<l (ii) b 2 >b> I. 

(2) IfZ>>0and£>0and 

b B' 

prove that ali < bA . Deduce that 

a a + A A 
~b < b + B B' 

(3) If a > ft and c > d, prove that a +c>b+d(i.e. inequalities can be 
added). If, also, b > and d > 0, prove that ac > bd (i.e. inequalities 
between positive numbers can be multiplied). 

(4) Show that each of the following inequalities may fail to hold even 
though a>Z>andc>d. 

(i)a-c>b-d 



Real numbers 



<>1 



(iii) ac>bd. 



What happens if we impose the extra condition that b > and d > 0? 

(5) Suppose that, for any e>0,a — e<b<a + e. Prove that a = b . 

(6) Suppose that a < b. Show that there exists a real number x satisfying 
a<x<b. 



1.9 Roots 

Let n be a natural number. The reader will be familiar with the notation 
y =x". For example, x 2 =x.x and* 3 = x.x.x. 

Our next assumption about the real number system is the following. Given 
any y>0 there is exactly one value of x > such that 

y = x". 

(Later on we shall see how this property may be deduced from the theory of 
continuous functions.) 

If y > 0, the value of x > which satisfies the equation y = x n is called the 
nth root of y and is denoted by 

x = y Un . 

When n = 2, we also use the notation y/y = y xn - Note that, with this con- 
vention, it is always true that y/y > 0. If y > 0, there are, of course, two num- 
bers whose square is y. The positive one is y/y and the negative one fc—y/y. 
The notation ± y/y means 'y/y or —y/y'. 

If r = m/n is a positive rational number and y > 0, we define 

y r = (y m y ,n . 

If r is a negative rational, then — r is a positive rational and hence y~ r is defined. 
If y > we can therefore define y r by 



Y = 



1 



We also write y° = 1 . With these conventions it follows that, if y > 0, then ^ r 
is defined for all rational numbers r. (The definition of y x when x is an irrational 
real number must wait until a later chapter.) 



1.10 Quadratic equations 

If j> > 0, the equation x 2 =y has two solutions. We denote the positive 
solution by y/y. The negative solution is therefore —y/y. We note again that 



Real numbers 
there is no ambiguity about these symbols and that ± y/y simply means 'y/y or 

-y/y. 

The genera] quadratic equation has the form 

ax 2 + bx + c = 

where a ¥= 0. Multiply through by 4a. We obtain 

4a 2 x 2 + Aabx + 4ac = 

{lax +b) 2 -b 2 +4ac = 

(2ax + b) 2 = b 2 -4ac. 

It follows that the quadratic equation has no real solutions if b 2 — 4ac < 0, 
one real solution if b 1 —4ac = and two real solutions if A 2 — 4ac > 0. If 
b 2 - 4ac > 0, 

2ax+b = ±y/(b 2 -4ac) 

-b ±y/(b 2 -4ac) 



x = 



2a 



The roots of the equation ax 2 + bx + c = are therefore 

-b-y/(b 2 -4ac) , . - b + y/(b 2 - 4ac) 

a = and u = ~ . 

2a v 2a 

It is a simple matter.to check that, for all values of x, 

ax 2 +bx + c = a(x-a)(x-p). 

With the help of this formula, we can sketch the graph of the equation 
y -ax 2 + bx + c. 



' y = ax 2 + bx +c 



V 



fl>0 



y = ax J - bx + e 



7^ 



a<0 



1.11 Example A nice application of the work on quadratic equations 
described above is the proof of the important Cauchy Schwarz inequality. This 
asserts that, if a,, a 2 , . . . ,a„ andZ>i,Z» 2 , . . . ,fe„ are any real numbers, then 



(a,6, +a 2 b? 



Bnh? < («i 



a\ 



... + a 2 n ){b 2 +b\ 



b 2 n ). 



8 



Real numbers 



Proof For any x, 

< (a l x + b l ) 2 + (a 2 x+b 2 ) 2 + ... + (a n x+b n ) 2 

= (a? + . . . + a 2 n )x 2 + 2ia.br + ... + a n h n )x + (b\ + . . . + b 2 n ) 

= Ax 2 + 2Bx + C. 

Since y = Ax 2 + 2Bx + O for all values of x, it follows that the equation 
Ax 2 + 2Bx + C = cannot have two (distinct) roots. Hence 

(2B) 2 -4AC<0 

i.e. B 2 <AC 

which is what we had to prove. 



1.12 Exercise 

(1) Suppose that n is an even natural number. Prove that the equation 
a-" = y has no solutions if y < 0, one solution if y = and two 
solutions if y > 0. 

Suppose that n is an odd natural number. Prove that the equation 
x" = y always has one and only one solution. 

Draw graphs of y = x 2 and y = x 3 to illustrate these results. 

(2) Simplify the following expressions: 

(OS 2 ' 3 (ii)27" 4 ' 3 (iii)32 6/s . 

(3) If y > 0, z > and r and s are any rational numbers, prove the follow- 
ing: 

(■) y r+s = ff (ii) y rs = (yj (iii) (yzj = fz\ 

(4) Suppose that a > and that a and are the roots of the quadratic 
equation ax 2 + bx + c = (in which b 2 — 4ac > 0). Prove that 

y = ax 2 + bx + c is negative when a < x < and positive when x < a 
orx >j3. Show also (without the use of calculus) that y = ax 2 + bx + c 
achieves a minimum value of c —b 2 /4a when* = —bj2a. 

(5) Leta,,a 2 , . . . , a„ be positive real numbers. Their arithmetic meanA n 
and harmonic mean //„ are defined by 






_ g,+g 2 + ...+a„ 



III. 1 



H-' = -- + -+... + - 
n\ai a 2 a„ 



Deduce from the Cauchy-Schwarz inequality that H n < A„. 
(6) Letai,a 2) ... ,a n andb i ,b 2 , ■ ■ ■ ,b n beany real numbers. Prove 
Minkowski's inequality, i.e. 



I (^ + b k y 



1/2 



< 



n 


1/2 


n 


>,al 


+ 


*:# 


k = i 




«t=i 



/?ea/ numbers i 

For the case « = 2 (or « = 3) this inequality amounts to the assertion 
that the length of one side of a triangle is less than or equal to the sum 
of the lengths of the other two sides. Explain this. 



1.13 Irrational numbers 

In § 1 .2 we mentioned the existence of irrational real numbers. That 
such numbers exist is by no means obvious. For example, one may imagine the 
process of marking all the rational numbers on a straight line. First one would 
mark the integers. Then one would move on to the multiples of j and then to 
the multiples of i and so on. Assuming that this program could ever be com- 
pleted, one might very well be forgiven for supposing that there would be no 
room left for any more points on the line. 



h tt i * 



iUtlLM 



-1 



But our assumption about the existence of nth roots renders this view unten- 
able. This assumption requires us to accept the existence of a positive real num- 
ber* (namely \J2) which satisfies x 2 = 2. If* were a rational number it would 
be expressible in the form 



m 

x = — 

n 



where m and n are natural numbers with no common divisor (other than 1). It 
follows that 



2 = In 2 



m 

and so m 2 is even. This implies that m is even. (If m were odd, we should have 
m = 2k + 1 . But then m 2 = 4k 2 + 4& + 1 which is odd.) We may therefore write 
m = 2k. Hence 

4k 2 = 2n 2 

n 2 = 2k 2 . 

Thus n is even. We have therefore shown that both m and n are divisible by 2. 
This is a contradiction and it follows that x cannot be rational, i.e. \/2 must be 
an irrational real number. 

Of course, \/2 is not the only irrational number and the ability to extract «th 
roots allows us to construct many others. But it should not be supposed that all 
irrational numbers can be obtained in this way. It is not even true that every 
irrational number is a root of an equation of the form 



10 Real numbers 

a + a x x + a 2 x 2 + . . . + a n x" = 

in which the coefficients a ,a t , ■ ■ ■ ,o n ate rational numbers. Real numbers 
which are not the roots of such an equation are called transcendental, notable 
examples being e and it. This fascinating topic, however, lies outside the scope 
of this book. 



1.14 Modulus 

Suppose that x is a real number. Its modulus (or absolute value) \x\ is 
defined by 

Ay 






1*1 = 



x (x>0) 
-x (x < 0). 




Thus |3 1 = 3, 1—41 = 6 and |0| = 0. Obviously |*| > for all values of*. 
It is sometimes useful to note that | x | = s/x 2 . 

1.15 Theorem For any real numbers:, 
-|x| <x<|x|. 

Proof Either x > or x < 0. In the first case, — |x| < < * = |x|. In 
the second case, — 1*| — X <0 < |*|. 

1.16 Tlieorem For any real numbers a and b 
\ab\ = \a\.\b\ 

Proof The most elegant proof is the following. 

\ab\ = s/((abf) = y/(a 2 b 2 ) = sj(a 2 ).yj(b 2 ) = \a\.\b\. 

The next theorem is of great importance and is usually called the triangle 
inequality. This nomenclature may be justified by observing that the theorem 
is the special case of Minkowski's inequality (exercise 1 .12(6)) with n = 1 . We 
give a separate proof. 

1.1 7 Theorem (triangle inequality ) For any real numbers a and b , 
\a + b\ < \a\+ |6|. 

Proof We have 



Real numbers 11 

\a+b\ 2 = {a + bf = a 2 + lab + b 2 

= \a\ 2 + 2ab + \b\ 2 

< \a\ 2 + 2\ab\+ \b\ 2 (theorem 1.15) 

= \a\ 2 + 2\a\-\b\ + \b\ 2 (theorem 1.16) 

= (\a\+\b\) 2 . 
It follows (from example 1 .6) that 

\a + b\ < |o|+ \b\. 

1.18 Theorem For any real numbers c and d, 

\c-d\ > |c|-|d|. 

Proof Take a = d and b = c — d in the triangle inequality. Then 
a + b =c and so 

|c| = \a + b\ < |a|+ \b\ = \d\ + \c-d\. 

1.19 Example 

(i) 2 = |-1+3|<|-1|+|3| = 1 +3=4.' 
(ii)7 = |6-(-l)|> |6|-|-1| = 6-1 =5. 

1.20 Exercise 

(1) Prove that |a| <b if and only if —b <a<b. 

[Recall that you have two things to prove as in example 1 .6.] 

(2) Prove that 

\c-d\ > \\c\-\d\\. 

(3) The distance d(x,y) between two real numbers x undy is defined by 
d(x,y) = |* —y\. Show that, for any x,y andz, 

(i) d(x,y)>0 (iii)d(;c,.y) = 0ifand only if x=y 

(ii) d(x,y)=d(y,x) (iv)d(x,y)<d(x,z) + d(z,y). 

1 (4) If r and s are rational numbers, prove that 

r + sy/2 

is irrational unless^ = 0. 
> (5) Suppose that the coefficients a¥=0,b and c of the quadratic equation 

ax 2 + bx+c = 

are all rational numbers and that a = r + s\/2 is a root of this equation, 
where r and s are also rational numbers. Prove that (3 = r —s\/2 is also a 
root, 
t (6) Prove that 3" 2 and 2 1/3 are irrational numbers. 



CONTINUUM PROPERTY 



2.1 Achilles and the tortoise 

The following is one of the famous paradoxes of Zeno. Achilles is to 
race a tortoise. Since Achilles runs faster than the tortoise, the tortoise is given a 
start of x feet. When Achilles reaches the point where the tortoise started, the 
tortoise will have advanced a bit, say x 1 feet. Achilles soon reaches the tortoise's 
new position, but, by then, the tortoise will have advanced a little bit more, say 
x 2 feet. This argument may be continued indefinitely and so Achilles can never 
catch the tortoise. 



A 


** 




-* 


v„ >• 

A 


m* 




■< 


x, >■ 

A - 



The simplest way to resolve this paradox is to say that Achilles catches the 
tortoise after he has run a distance of x feet, where x is 'the smallest real number 
larger than all of the numbers x ,x + x u x + x y x 2 , . . .'. Zeno's argument 
then simply reduces to subdividing a line segment of length x into an infinite 
number of smaller line segments of respective lengths x ,x lt x 2 , . . . 




Formulated in this way, the paradox loses its sting. 

This solution, of course, depends very strongly on the existence of the real 
number x, i.e. the smallest real number larger than all of the numbers x , 
x +*!, jc +x$ +x 2 , . ■ ■ 



12 






Continuum property 13 

A similar lesson is to be learned from Archimedes' method of exhaustion. 
This was invented by Archimedes as a means of evaluating the area of regions 
with curved boundaries. 

In the case of a circle, the method involves inscribing a sequence of larger 
and larger regular polygons inside the circle and then calculating their areas 
Ai,A 2 ,A 3 , . . . 




The area A of the circle is then identified with the 'smallest real number 
larger than each of the numbers/^, A 2 ,A 3 , . . .'. By taking the radius of the 
circle to be 1 and calculating the value of A n for a large enough value of n, 
Archimedes was able to obtain as close an approximation to -n as he chose. 

Note that this method again depends strongly on the existence of a smallest 
real number larger than all of the real numbers Ai,A 2 ,A-$,... 



2.2 The Continuum Property 

The discussion of the previous section indicates the necessity of making 
a further assumption about the real number system. It would, after all, be most 
unsatisfactory if Achilles never caught the tortoise or if a circle had no area. We 
shall call our new assumption the Continuum Property of the real numbers. 

The Continuum Property is not usually mentioned explicitly in a school 
algebra course, but it is very important for what follows in these notes. The rest 
of this chapter is therefore devoted to its consideration. We begin with some 
terminology. 

A set S of real numbers is bounded above if there exists a real number H 
which is greater than or equal to every element of the set, i.e. if, for some //, 



x<H 



for any x e S. 

The number H (if such a number exists) is called an upper bound of the set S. 
A set S of real numbers is bounded below if there exists a real number h 



14 Continuum property 

which is less than or equal to every element of the set, i.e. if, for some h. 



Continuum property 



B<H. 



15 



i 



B \H 



x>h 



for any xGS. 

The number h (if such a number exists) is called a lower bound of the set S. 

A set which is both bounded above and bounded below is just said to be 
bounded. 

2.3 Proposition A set S of real numbers is bounded if and only if there 
exists a real number K such that 

|jc| < K 

for any xG S. (This is easily proved with the help of exercise 1 .20(1).) 



2.4 Examples 

(i) The set {1 , 2, 3} is bounded above. Some upper bounds are 100, 10, 
4 and 3. The set is also bounded below. Some lower bounds are —27, and 1 . 

(ii) The set {x : 1 < x < 2} is bounded above. Some upper bounds are 100, 
10, 4 and 2. The set is also bounded below. Some lower bounds are —27, and 
1. 

(iii) The set {x : x > 0} is unbounded above. If H > is proposed as an upper 
bound, one has only to point to H + 1 to obtain an element of the set larger 
than the supposed upper bound. However, the set {x :x> 0} is bounded below. 
Some lower bounds are —27 and 0. 



We now state the Continuum Property which will be fundamental for the 
remainder of these notes. 



Continuum Property 

Every non-empty set of real numbers which is bounded above has a 
smallest upper bound. Every non-empty set of real numbers which is 
bounded below has a largest lower bound. 



Thus, if S is a non-empty set which is bounded above, then S has an upper 
bound B such that, given any other upper bound //of S, 



Similarly, if S is a non-empty set which is bounded below, then S has a lower 
bound b such that, given any other lower bound h, 



b>h. 






2.5 Examples 

(i) The smallest upper bound of the set {1,2, 3} is 3. The largest lower 
bound of the set {1,2, 3} is 1. 

(ii) The smallest upper bound of the set {x : 1 < x < 2} is 2. The largest lower 
bound is 1 . 

(iii) The set {x : x > 0} has no upper bounds at all. The largest lower bound 
of the set {x:x>0}is0. 



2.6 Supremum and infimum 

If a non-empty set S is bounded above, then, by the Continuum 
Property, it has a smallest upper bound B. This smallest upper bound B is some- 
times called the supremum of the set S. We write B — sup 5 or 

B = sup x. 

i6=S 

Similarly, a set which is bounded below has a largest lower bound b. We call 
b the infimum of the set S and write b = inf 5 or 

b = inf x. 

xGS 

Sometimes you may encounter the notation sup S = + °°. This simply means 
that 5 is unbounded above. Similarly, inf 5 = — °° means that S is unbounded 
below. 



2.7 Maximum and minimum 

If a set S has a largest element M, we call M the maximum of the set S 
and write M = max S. If 5 has a smallest element m, we call m the minimum of 
S 1 and write m = min S. 



16 



Continuum property 



It is fairly obvious that, if a set S has a maximum M, then it is bounded above 
and its smallest upper bound is M. Tlius, in this case, sup S = max S. 

A common error is to suppose that the smallest upper bound of a set S is 
always the maximum of the set. However, some sets which are bounded above 
(and hence have a smallest upper bound) do not have a maximum. (See example 

2.8(a).) 

Similar remarks, of course, apply to largest lower bounds and minima of sets. 



2.8 Examples 

(i) The set {l , 2, 3} has a maximum 3 and this is equal to its smallest 
upper bound. The set {l , 2, 3} has minimum 1 and this is equal to its largest 
lower bound. 

(ii) The set {x : 1 < x < 2} has no maximum. The number 2 cannot be the 
largest element of the set because it does not belong to the set. On the other 
hand, any x in the set satisfies 1 < x < 2. But then 



x + 2 



y = 



is an element of the set which is larger than x. Hence x cannot be the largest 
element of the set. However, {x : 1 < x < 2} has smallest upper bound 2. 

The set {x : 1 < x < 2} has minimum 1 (why?). This is equal to its largest 
lower bound. 

(iii) The set {x : x > 0} has no maximum, nor does it have any upper bounds. 
The set {x : x > 0} has no minimum (why not?). Its largest lower bound is 0. 



2.9 



Intervals 



An interval I is a set of real numbers with the property that, if x E I 
and y Eland x<z^y, then zEI, i.e. if two numbers belong to /, then so does 
every number between them. 

In describing intervals we use the following notation: 



(a, b) = {x:a<x<b} 
[a,b] = {x:a<x<b} 
[a,b) = {x:a<x<b} 
(a,b] = {x:a<x<b}. 



a b 



wmm 



(a,.' ) 









T 



Continuum property 



17 



These are the bounded intervals (classified by whether or not they have a 
maximum and whether or not they have a minimum). We also need to consider 
the unbounded intervals. For these we use the notation below. 



(a, °°) = {x : x >a} 

[a,°°) = {x :x>d) 

(-°°,b) = {x:x<b} 

(-°°,b] = {x:x<b} 



V7/////////'y/7//7/7////y, 



(«,») 



■/y////////////7\ 

W //W////7W 



(Do not be misled by this notation into supposing that °° or — °° are real num- 
bers or that they can be treated as such.) 

All intervals (with the exception of the empty set and the set of all real 
numbers) fall into one of the categories described above. We call the intervals 
(a,6),(a,°°)and(— °°,b)open and the intervals [a, b], [a, °°) and (— °°,fe] 
closed. (The intervals [a, b) and (6, a] are sometimes called half-open.) 

Of particular importance are the closed, bounded intervals [a, b] . We shall 
call such an interval compact. 



2.10 Exercise 

(1) Which of the following statements are true? 

0)36(1,2) (ii)2e(-°°,3] (iii) 16 [1,2) 

(iv)2e(0,2) (v)3G[l,3]. 

(2) If % is a real number and 5 > 0, prove that 

{x:|£-x|<S} = a-S,£ + 5). 

(3) Decide in each of the following cases whether or not the given set is 
bounded above. For those sets which are bounded above, write down 
three different upper bounds including the smallest upper bound. 
Decide which of the sets have a maximum and, where this exists, write 
down its value. 

(i)(0,l) (ii)(-°°,2] (iii) {-1,0, 2, 5} 

(iv)(3,°°) (v) L0, 1]. 

(4) Repeat the above question but with the words 'above, upper and maxi- 
mum' replaced by 'below, lower and minimum'. 

(5) Give an example of a set which has smallest upper bound 4 but contains 
no element x satisfying 3 < x < 4. 

(6) Show that, given any element x of the set (0, °°), there is another 
element y 6 (0, °°) with the property that ;' < x. Deduce that (0, °°) 
has no minimum. 



18 



Continuum property 



Continuum property 



19 



2.11 Manipulations with sup and inf 

We prove one result and list some others as exercises. 

2.12 Theorem Suppose that S is a non-empty set of real numbers which is 
bounded above and % > 0. Then 

sup %x = % sup x. 

Proof Let B = sup S. Then B is the smallest number such that, for any 
x<ES, 

x<B. 

Let 2"- {#* : *€ S}. Since | > 0, 

for any* G 5. Hence Tis bounded above by £/?. By the Continuum Property, 
rhas a smallest upper bound (or supremum) C. We have to prove that C = %B. 
Since %B is an upper bound for 7* and C is the smallest upper bound for 7 1 , 

Now repeat the argument with the roles of S and T reversed. We know that C 
is the smallest number such that, for any y&T, 

y<C. 

Since £ > 0, it follows that 

for any y G T. But S = {|"V : ? G 7% Hence ?"' C is an upper bound for S. But 
# is the smallest upper bound for S. Thus 

£5<C. 
We have shown that C< |5 and also that £fi < C. Hence \B = C. 



2.13 Exercise 

(1) Suppose that S is bounded above and that S C S. Prove that 
sup S < sup 5. 

(2) Suppose that S is bounded above and that £ is any real number. Prove 
that 

sup (x + £) = sup x + %. 

*£S i£S 



(3) Suppose that 5 is bounded above. Prove that 

inf (— x) = — sup x. 
i£s .*es 

Hence obtain analogies of theorem 2.12 and exercises 2.13(1 and 2) in 
which inf replaces sup. 

(4) The distance d{%, S) between a real number £ and a non-empty set 5 of 
real numbers is defined by 

d(S,S) - inf |$-*| 

i£S 

(see exercise 1.20(3)). Find the distance between the real number % = 3 
and the sets 

(i)S = {0, 1,2} (ii)5 = (0,l) 

(iii)S=[l,2] (iv)S = (2,3). 

(5) (i) If £ G S, prove that d(|, 5) = 0. Give an example of a real num- 

ber ? and a non-empty set 5 for which d(£, S) = but £ G" S. 
(ii) If 5 is bounded above and % = sup S, prove that d(|, 5) = 0. 

Deduce that the same is true if 5 is bounded below and £ = inf S. 
"I" (iii) If /is a closed interval, prove that d{%, /) = implies that \ G /. 

If/ is an open interval (other than R or 0), show that a j- ^/can 

always be found for which d(£, /) = 0. 
f (6) Suppose that every element of an interval / belongs to one or the other 
of two non-empty subsets 5 and T. Show that one of the two sets S or 
T contains an element at zero distance from the other. (Do not assume 
that S and Tare necessarily intervals. The set S, for example, could 
consist of all rational numbers in / and the set T of all irrational num- 
bers in /.) 

[Hint : Suppose that $ < t, where s G S and t G T. Let T = {x : x G T 
and x > s}. Show that To ^ and is bounded below. Write b = inf T 
and consider the two cases b € Tand b ^ T.\ 






NATURAL NUMBERS 



3.1 Introduction 

In this chapter we explore some implications of the Continuum 
Property for sets of natural numbers. As always, arithmetical properties are 
taken for granted. But where a proof hinges on such a property, the property 
will be explicitly stated. 



3.2 Archimedean property 

The Archimedean property is the assertion of the following theorem. 

3.3 Theorem The set M of natural numbers is unbounded above. 

Proof For this proof we need to assume that, if n is a natural number, 
then so is n + 1 . 

Suppose that the theorem is false. Then N is bounded above. By the Con- 
tinuum Property it therefore has a smallest upper bound B. Since B is the 
smallest upper bound for N, B — 1 is not an upper bound for N. Thus, for some 
«€N, 

n >B-\ 

n+ I >B. 

But then n + 1 is an element of the set fsj which is greater than B. But B is an 
upper bound of the set M. This is a contradiction. 
Hence M is unbounded above. 



3.4 Example Prove that the set 

5 = {«-' :neN} 

is bounded below with largest lower bound 0. 

Proof Since the natural numbers are all positive, 

o 
n _1 >0 



20 



Natural numbers 



21 



for all n € foJ. Hence the set S is bounded below by 0. We have to prove that is 
the largest lower bound . 

Suppose that /; > is a lower bound for S. Then, for each n € M, 

n' 1 >h. 

Hence, for each n e fol , 

n « ft" 1 . 

But then h' 1 is an upper bound for the set M which we know is unbounded 
above. It follows from this contradiction that no h > can be a lower bound for 
S. Thus is the largest lower bound for S. 

Note that 0^5 and hence S has no minimum. 



3.5 Theorem Every set of natural numbers which is not empty has a mini- 

mum. 

Proof For this proof we need to assume that the distance between 
distinct natural numbers is at least 1 . 

Let S be a non-empty set of natural numbers. Given that all natural numbers 
are positive, is a lower bound for S. By the Continuum Property, it follows 
that S has a largest lower bound b. Since b is the largest lower bound, b + 1 is 
not a lower bound. Therefore, for some «SS, 

n < b + 1 . 

If n is the minimum of S, there is nothing to prove. If not, then, for some 
wj e S,m< n. We obtain the inequality 

b < m < n < b + 1 

from which follows the contradiction < n — m < 1. 



3.6 Exercise 

(1) Prove that the set 

n—l 



s = 



:«e 



is bounded above with smallest upper bound 1 . Does S have a maxi- 
mum? 
(2) Let X > 1 . Prove that the set 

S = {X n :n&U} 

is unbounded above. [Hint: if B were the smallest upper bound, then 
BX~ l could not be an upper bound.] Show that, if 0<x < 1, then the 



22 



Natural numbers 



set 

T = {x n :nGN} 

is bounded below with largest lower bound 0. 

(3) Prove that any non-empty set of integers which is bounded above has a 
maximum and that any non-empty set of integers which is bounded 
below has a minimum. 

(4) Let a < b. Prove that there exists a rational number r satisfying 
a<r<b. [Hint: justify the existence of a natural number n satisfying 
n~>(b —a)~ l and consider the rational number r = tn/n where m is the 
smallest integer satisfying m>an.] 

(5) Let S be the set of all rational numbers r which satisfy < r < 1 . Show 
that S has no maximum and no minimum. Prove that S is bounded and 
has largest lower bound and smallest upper bound 1 . [Hint: for the 
last part, use the previous question.] 

(6) Let a < b. Prove that there exists an irrational number £ satisfying 
a < % < b. [Hint: exercise 1.20(4).] 



3.7 Principle of induction 

Suppose that a line of dominoes is arranged so that, if the «th one falls, 
it will knock over the (n + l)th. If the first domino is pushed over, most people 
would agree that all the dominoes would then fall down. 

The principle of induction is an idealisation of this simple notion. We show 
below how it may be deduced from theorem 3.5. 

3.8 Theorem (principle of induction) 

Suppose that, for each n £ M , P(n) is a statement about the natural 
number n. Suppose also that 
(i)P(l)is true 

(ii) if P(n) is true, then P(n + 1) is true. 
Then P(n) is true for every n £ftj . 

Proof For this proof we need to assume that, if n is a natural number 
other than 1 , then « — 1 is a natural number as well. 

Let S = {« : P{n) is false}. We want to prove that S is empty. Suppose that S 
is not empty. Then from theorem 3.5 it follows that S has a minimum m. Since 
P(l) is true, 1 (fc S. Hence m =fc 1 . Therefore m — \ is a natural number. But m is 
the minimum of S. Thus m — 1 S and so P(m — 1) is true. But then P(m) is 
true (because of hypothesis (ii) of the theorem). Hence m $ S which is a contra- 
diction. 



Natural numbers 23 

3.9 Example For each n£N 

1 +2 + 3 + . . .+« = \n(n + 1). 

Proof Let S„ = 1 +2 + 3 + ... + «= 2£ =1 k. Let P(n) be the state- 
ment S n = \n{n + 1). 

We have S 1 = l,and \. 1(1 + 1) = 1. Hence P(l) is true. We now wish to 
show that if P{n) is true, then P(n + 1) is true'. To do this, we assume that P(ri) 
is true and try and deduce that P{n + 1) is true. Thus we assume that S„ = 
\n(n + 1) and try and deduce that 5 n+1 = k(n + 1)(" + 2). But 

S n + l = 1 + 2 + 3 + . . . + n + (n + 1) 

= S n + (n + l) 

= \n(n + 1) + (« + 1) 

= &« + l)(n + 2) 

as required. The result now follows by induction. 



3.10 Example The real number 

A n = —(at +a 2 + ■ • - + a n ) 
n 

is the arithmetic mean of the numbers a 1 ,a 2 , ... ,a„. If 01,22, ... ,a n are 
positive, their geometric mean G n is defined by 

G n = (a,a 2 ...fl n ) 1/n . 

The 'inequality of the arithmetic and geometric means' asserts that 

G n <A n . 

Proof This inequality has an entertaining proof using 'backwards 
induction' (see exercise 3.1 1(5)). 

LslP(n)be the statement 'For any positive numbers at, a 2 , . . . ,a n , 
G„ < A„'. We shall show that 

(i) P(2 n ) is true for each nGN. 

(ii) If P(n) is true, then P(n — 1) is true. 
The result then follows by backwards induction. 

(i) We begin by proving P(2). Now, 

< (V^i - y/a 2 ) 2 -at — 2yJ(a x a 2 ) + a 2 

( fl|a2 ) 1/2 <K«i+^)- 

This is simply P(2). We now show that P(2 n ) implies P(2" * '). Let m = 2". Then 
2«+i _ 2m. We have to assume P(m) and try and deduce P(2m). Since P(m) is 
true, 



24 Natural numbers 



(a 1 a 2 ...a m ) l " n <-(a 1 +a 2 



1 



•o 



and ( a m*\ a m+2 • ■ - a 2m) < ~~ ( a m + 1 + "m + 2 + • ■ • + «2m)- 

But we know that P(2) is true. Hence 

{( flia2 ...0 ,/m (^^m + 2---«2 m ) 1/m }" 2 

^ 1 (a, + . . . + a m a m+1 +...+a 2m 



■ + 



m 



m 



Thus 



( ai a 2 . ..a 2m ) V2m < — (a, + <z 2 4- . . . +a 2m ) 
2ffi 



which is /'(2/m). 

We have shown that P(2 1 ) is true and that, if P(2 n ) is true, then P(2 n+1 ) is 
true. Hence P(2") is true for all n S |^J by straightforward induction. 

(ii) We now show that, if P(n) is true, then P(n — 1) is true. Suppose that 
P(n) is true. Then 



{(ha i ...a n . t G n - l y n < 



vn ^ai+a 2 + ... + a„_, + G„_i 



-»-] r MM / (H-lMn-l+Cn-1 



ft 

G n _, </!„.,. 

ThusP(n — l)is true. 

This completes the proof. 

3.11 Exercise 

(1) Prove by induction that 

n 



(i) I & 2 = l 2 + 2 2 + . . . + n 2 = 



fe = i 



(ii) T A 3 = l 3 + 2 3 + . . . + n 3 
(2) Prove by induction that, if x =£ 1, 



£* fe = 1+x+x 2 + ...+*" 
fe=o 



in(n + l)(2n+I) 

i« 2 (« + l) 2 . 

l-x n+1 



1-as 



Natural numbers 



25 



Deduce that, for any real numbers a and b, 

a"-b" m (a-bXa"' 1 + a"' 2 b + a"' 3 b 2 + . . .+ ab"' 2 + b"' 1 ). 

(3) A polynomial P(x) is an expression of the form 

P(x) = a n x n + a^iX"' 1 + . . . + aix+a . 

If a n =£ 0, the polynomial is said to be of degree n. 

Suppose that P(x) is a polynomial of degree n and that, for some if, 
P{£) = 0. Prove that P(x) = (x - %)Q(x) where Q(x) is a polynomial of 
degree n — 1 . [Hint: use the previous question.] If gj, 2- 2 , . . . , £„ are 
distinct real numbers and P{%\) = P(£ 2 ) = . . . = P(% n ) = 0, prove that 

P(x) = k(x-^)(x-$ 2 )...(x-U 

where A: is a constant. 

(4) Assuming that w! = 1.2.3...n and 0! = 1, we define 



m 



r\(n-r)\ 
Prove that 



(r=0,l,2,...,ii). 



«+l 

r 



(r=l,2,...,n). 



Using this result and the principle of induction, obtain the binomial 
theorem in the form 

{a + bf = a" + na"' 1 b + ^^ a"- 2 b 2 + . . . + b" 



= y. 



V 



a"- r b r . 



(5) Let P(n) be a statement about the natural number n and suppose that 
(i) P(2") is true for each n G M 

(ii) if P(n) is true, then P(n — 1) is true. 

Prove that P(n) is true for every n€N. [Hint: the set {2" : n S M } is 

unbounded above (exercise 3. 6(2)). J 

(6) Suppose that < X< Y and that x and y belong to the interval [X, Y]. 
Prove that, for each nGM, 

AT"" | jc -y K «^TK| x 1/n -/'" | < YX 1 '" \x-y\. 

[Hint: exercise 3.1 1(2).] 



Convergent sequences 



27 



= !{1 + (" 



£)+(-£)*+... + (-|f-*} 



CONVERGENT SEQUENCES 



4.1 The bulldozers and the bee 

Two bulldozers are moving towards each other at a speed of one mile 
per hour on a collision course. When they started they were one mile apart and a 
bee was perched on the front of one of them. The bee began to fly back and for- 
ward between the bulldozers at a constant speed of two miles per hour, vainly 
seeking to avoid his fate. How far from his starting point will the unfortunate 
insect be crushed? 



fef" 




fe# fr^S 



This riddle is usually posed in the hope that the victim will embark on some 
complicated calculations involving the flight of the bee. But it is quite obvious 
that the bee will be crushed when the bulldozers collide. Since the bulldozers 
travel at the same speed, this will happen halfway between their starting points. 
The answer is therefore one half mile. 

Let us, however, take up the role of the riddler's victim and examine the 
flight of the bee. Letx„ denote the distance the bee is from his starting point 
when he makes his n th landing. Then 

r _ 2 
x \ — 3 

v - 2_2 

x 2 — 3 9 



and, in general. 



= §-!+f 7 -... + (-iy- 1 2(i)" 



26 



2 li -(-!)" 



3 ll _ ( _, )1 =^a-(-r). 

How is the answer \ to be extracted from this formula? 

It seems clear that, as n gets larger and larger, (— £)" gets closer and closer to 
zero and so x n gets closer and closer to \. We say that 'x n tends to § as n tends to 
infinity' - or, in symbols, l x n -> \ as n -*■ °°\ 

This idea is of the greatest importance, but, before we can make proper use of 
it, it is necessary to give a precise formulation of what it means to say that x n -> I 
as n -* °°. 



4.2 Sequences 

A sequence may be regarded as a list of numbers Xi,X2,x 3 ,x 4 , . . . 
More precisely, we can say that a sequence is determined by a rule which assigns 
to each natural number n a unique real number x n . We call x n the n th term of 
the sequence. 
The notation 

means the sequence whose «th term isx„. 

The set {x n : n € N} is called the range of the sequence. We say that a 
sequence is bounded above (or below) if its range is bounded above (or below). 
Thus (x„) is bounded above by //if and only if-xr„ <//(« = 1, 2, 3, . . .). 



4.3 



Examples 

(i)<(-ir> = -i,+ i,-i, + i,... oo<«- 1 >= i,|, M.. 

(iii)<l> = 1,1,1,1,... (iv)(2 n > = 2,4,8,16, 

(v) The sequence (x„) defined inductively by *i = 2 and 
1 



(n = 2,3,4, ...). 
The first few terms are 2, i. 12. I§> • ■ • 



X "~ 2Y n - l+ x n . r 



4.4 Definition of convergence 

A sequence (x n ) is said to converge to the limit I if and only if the 



following criterion is satisfied. 



28 



Convergent sequences 



Given any e > 0, we can find an N such that, for any n>N, 
\x n -l\<e. 



We write x n ■* I as n ■* °° or 
lim x„ = /. 



It may be helpful to think o{x h x 2 ,x 3 , ... as successive approximations to 
the number /. The distance \x n —l\ between x n and / is then the error involved in 
approximating / by x„. The definition of convergence then simply asserts that we 
can make this error as small as we choose by taking n large enough. 

The diagram is supposed to represent a sequence <x„> with the property that 
x„ -*■ I as n -*• <*>. 




n>6 

For the value of e indicated in the diagram, a suitable value for N in the 
definition isN= 6. For each value of« >6, 

\x n -l\<e. 

In particular, the value of \x 9 — l\ has been noted in the diagram. 

Notice that the definition of convergence begins 'Given any e > 0, we can 
find an N . . . ' Here the emphasis is on the very small values of e > 0. It is clear 
from the diagram above that, if we had chosen to look at a very much smaller 
value of e > 0, then we should have had to have picked out a very much larger 
value of N. In general, the smaller the value of e > 0, the bigger must be the 
corresponding value of iV. 



4.5 Example 
1 



1 + — -* 1 as n -*■ °°. 
n 



Proof Let e > be given. We must find a value of ./V such that, for any 



n>N, 



Convergent sequences 



29 



But 



1 »'" 



«'" 



<e. 



and so we simply need to find an N such that, for any n>N, 



i.e. 



-<6 

n 



e 



But this solves the problem. We simply chose A' = 1/e. Then, for any n>N, 



i.e. 



i.e. 



n>N = - 
e 



-<e 

n 



l+ p. 



<e. 



We have shown that, given any e > 0, we can find an N (namely N=l/e) 
such that, for any n>N, 






<e. 



It is important to observe that N depends on e, i.e. for each e > we use a 
different N. Some values of e and the corresponding values of N are entered in 
the table below. 



e 


N 


0-4 


2-5 


0-1 


10 


0000 001 


1 000 000 



(Note: Some authors insist that A' be a natural number. This makes the defi- 
nition of convergence a little more elegant but renders examples like that above 
marginally more complicated. If we wanted A' to be a natural number in the 
example above, we could not simply write A' = 1/e. Instead we should have to 
choose N to be some natural number larger than 1/e.) 



30 Convergent sequences 

4.6 Exercise 

(1) Prove that 

r-fc4i = 



n-"~ \n 2 + 1 , 

(2) Let r be any positive rational number. Prove that 

— -* as n •* °°. 
n 

(3) Let X be any real number. If x n -* I as n -> °°, prove that Xx n -> X/ as 



4.7 Criteria for convergence 

In the example and exercises above it is fairly easy to decide what value 
of A' is appropriate to any given value of e > 0. But this is by no means always 
the case. It is therefore natural to look around for some shortcuts which will 
enable us to determine whether a sequence converges (and what its limit is) with- 
out our having to indulge in the painful process of appealing to the definition. 
In this and the next section we give some useful results of this sort. 

4. 8 Proposition (combination theorem) Let x„ ■* / as n -* °° and y n -*m as 
n •* °° and let X and /j be any real numbers. Then 

(i) Xx n + ny n -*Xl + [xm as n -* °° 

(ii) x n y n -* Im as n -* °° 

x I 

(iii) — -*■ — as n ■* °° (provided that in i= 0). 
y n rn 

The proofs of (ii) and (iii) can be found in the appendix. These proofs are 
somewhat technical and we prefer not to hold up the discussion by presenting 
them at this stage. Readers who prefer to omit the proofs of (ii) and (iii) 
altogether will not suffer greatly as these results may be deduced from (i) once 
the theory of the exponential and logarithm functions has been developed. 

Proof of 4.8(ij. After exercise 4.6(3) we need only show that, if x n -* I 
aSH -+ °° and y n -*m asn-*°°, thenx n + y n ->l + m asn -»°°. 

Let e > be given. Then |e > 0. Since x n -* I as n ■* °°, it follows that we can 
find an A' 1 such that, for any n >N U 

\x n -l\<\e. (1) 

Similarly, we can find an A^ such that, for any n>N 2 , 

\y n -m\<\e. (2) 



Convergent sequences 



31 



Let Af be whichever of N t and Af 2 is the larger, i.e. A' = max {A^, A^}- Then, if 
n >N, both inequalities (1) and (2) are true simultaneously. Thus, for any 
n>N, 

\(x„+yn) -(/ + «)! = \(x n -l) + (y n -m)\ 

< \x n —l\ + \y n —m\ (triangle inequality) 

<\e + \e = e. 

Given any e > we have found a value of N (namely N = max {N\, N 2 }) such 
that, for any n > N, \(x n +y„) — (I + m)\< e. Hence x n +j„-»; + masn^». 



4.9 Example Prove that 

/ 2n 3 - 3ra 
.»?- ( 5 „3 + 4« 2 -2 

Proof We write 

2n 3 -2n 2-3/T 2 



2-0 2 

= — as n -* °°. 



5n 3 + 4n 2 -2 5 + 4«-'-2« -3 5 + 0-0 5 

The supporting argument is as follows. It is obvious from the definition of con- 
vergence that 2 -* 2 as n •* °°. From exercise 4.6(2), n' 2 -*■ as n •* °°. Hence, 
by proposition 4.8(i), 

2-3w" 2 ^-2-3.0 = 2as«-"°°. 

Similarly, 

5+4n- 1 -2M- 3 ^-5+4.0-2.0 = 5 as «->">=. 

The result then follows from proposition 4.8(iii). 

4.10 Theorem (the sandwich theorem) Suppose that y n •* I as n •* °° and 
thatz n -»Z as«-*°°. Uy n <x n <z n (n = 1, 2, . . .), then 

x„ -* / as n ■* °°. 

(Here the sequence <^: n > is 'sandwiched' between the two sequences <>> n > and <z n > 
- just as the bee of §4.1 was 'sandwiched' between the bulldozers.) 

Proof For this proof it is necessary to note that the inequality 
\x — 1\ <eis true if and only if/ — e<x</+e. (See exercise 1.20(1).) 



l-e 



i 
->♦«- 



/ + , 



7£ 



I.V-/I 



32 



Convergent sequences 



Let e > be given. We have to find an N such that, for any n>N, 
\x n -l\<e. 

Since y n -* I as n -> °°, we know that there exists an N t such that, for any 
n>N u 

\)' n -I\<e. (3) 

Similarly, since z„ -* / as n -* °°, there exists an N 2 such that, for any n >N 2 , 

\z n -l\<e. (4) 

Let A' be whichever of N t and A^ is the larger, i.e. TV = max {JVj.A'j}. Then, if 
n >N, both inequalities (3) and (4) are true simultaneously. Thus, for any 
n>N, 

l-e<y n <l + e 

and / — e<z n </ + e. 

But_y„ <x„ <z„ (n = 1, 2, . . .). Hence, for any n >N, 

!-e<y n <x n ^z n <l + e. 

Hence / — e<x n <l + e 

i.e. \x„—l\<e. 

Given any e > 0, we have found a value of A' (namely JV = max {N%, N 2 }) such 
that, for any n>N, 

\x n -l\<e. 

Hence x n -*■ I as n -* °°. 

4.11 Corollary Suppose that v„ -» as n -> °° and that 

\x n -l\<y n («=l,2,...). 
Then x n -* I as n ■+ °°. 

Proof The inequality \x n — 1\ < y n is equivalent to / — y n < x n < / + y n . 
But, using proposition 4.8, / — y n -> / as n -» 0° and / + _y„ ■+ / as « -»«>. Hence 
x n -> / as n ■* °°, by theorem 4.10. 



4.72 



Example Suppose that |.v| < 1. Prove that 
x" -»• as « -» °°. 
TVoo/ Write 

. .. 1 1 



xr = 



l 
< — . 



Thus 



(!+/?)" 1 + nh + * 2 n(n - \)h 2 + . . .+h" nh 



Convergent sequences 



33 



\x n -Q\ = \x\ n <— («= 1,2,.. .). 

But 1/rt -> as « -*■ °° (by exercise 4.6(2)). Thus, by corollary 4.1 1 , 
x" •* as n -*■ °°. 



4.13 Example In the problem of the bulldozers and the bee, we obtained 
the sequence 

By example 4. 1 2, (— J)" -> as « -*■ °° and hence x„ -» \ as w •* °°. 

4.74 Example Let x > 0. Prove that 

x m ■* 1 as n -> °°. 

/>oo/ In the inequality of the arithmetic and geometric means, take 
1 = a , = fl:2 = . . . = a n _, and a„ = .*. Then G„ = x 1/n and>4 n = (n — 1 + *)/« 
(x-l)/n + 1. 

y — 1 

Hence x 1,n < +1 



i.e. 



: i/"_! < _A ( _iy 



(Alternatively, take X =y = 1 in exercise 3.1 1(6).) 
Assume to begin with that x > 1 . Then 

0<x 1/n -l <-(x-l) 

and hence *"" -* 1 as n ■+ °° by the sandwich theorem. If < x < 1 , then 
x =y~ l where y > 1. Buty 1 '"-*- 1 as w -► o °and hence 

1 1 
by proposition 4.8(iii). 



4.15 Monotone sequences 

A sequence Cc„> is increasing if 
x n+l >x„ (n = l,2, ...). 



34 



Convergent sequences 



Similarly, <x n > is decreasing if 

X n+l <X„ (rt=l,2 ). 





(.v., > increases 



(v„ ) decreases 



A sequence Oc„> is strictly increasing if x n + 1 > x n (n = 1 , 2, . . .). It is strictly 
decreasing if x n + l < x n (n = 1 , 2, . . .). 

A sequence which is either increasing or else decreasing is called monotone. 



4.16 Example 

(i) The sequence (2"> is strictly increasing. 

(ii) The sequences (n~ l ) and (— n) are both strictly decreasing. 

(iii) The sequence (1) is both increasing and decreasing. 

(iv) The sequence {(— l) n > is not monotone. It is neither increasing nor 
decreasing. 



4.17 Theorem 

(i) If <.x„) is increasing and bounded above, then it converges to its 
smallest upper bound. 

(ii) If <*„> is decreasing and bounded below, then it converges to its largest 
lower bound. 

Proof If (x„) is decreasing and bounded below, then <— x n ) is increasing 
and bounded above. Hence it is only necessary to prove (i). 

Suppose that (,x n ) increases and is bounded above with smallest upper bound 
B. We must show that x n -*■ B as n -*■ °°. 

Let e > be given. Since B — e is not an upper bound, there exists at least 
one term x N of the sequence such that 

x N >B-e. 

But (x„) increases. Hence, for any n >N,x„ ~>x^. Thus, for any n>N, 
x„>x N >B — e. 



Convergent sequences *S 

Also B is an upper bound for the sequence. Hence x„ < B (n = 1 , 2, . . .). 
Thus, for any n>/V, 

B-e<x n <B. 

B-e<x n <B + e 

i.e. \x n -B\<e. 

Given any e>0, we have found an Af such that, for any n >N, \x n — B\<e. 
Hence x n -* B as n -*■ <*>. 




4.18 Examples (i) The sequence <(« - l)/«) increases and we know from 
exercise 3.6(1) that it is bounded above with smallest upper bound 1 . By 
theorem 4.17 it follows that 



«-l 



1 as n -*■ <*>. 



(ii) If < x < 1 , the sequence Ct"> decreases and is bounded below with 
largest lower bound (exercise 3.6(2)). Thus 



as n 



4.19 Example The examples above are nothing new. But consider the 
sequence 

((] +«-')">. 

We first show that this sequence increases. In the inequality for the geometric 
and arithmetic means, take a^ — a 2 = ■ ■ ■ — a n -\ = 1 + (« — 0" and a n = 1 . 
Then 



36 



Convergent sequences 



Hence 



x | i y"- i ^ < (^-ixi + (/2 -in + 1 = [j + i 



1 +■ 



n-1 



<ll+4l §1-2,3,...) 



and so the sequence increases. To be able to deduce the convergence of the 
sequence from theorem 4.17, we also need to prove that the sequence is 
bounded above . By the binomial theorem, 

'l\ n(n-\)(\ U 



,+ b ■"•l? 



SI- M; 



: ■ -2i-h+ H H)h- 



" \ nl 



<I + 1 U + ±+...+l 

2! 3! r! 



1 - 



«-l\ 1 



n I n\ 



1 1 1 

< 1 + 1 +-+-^+ .. . + -^r, (because 2" _1 <«!) 

= 1+: tz? = i +2(1- an < 3. 

Hence the sequence is bounded above by 3. From theorem 4.17 it follows 
that ((I + aT 1 )"} converges to a limit B and B < 3. (In fact, B = e = 2-718 . . .) 



4.20 Exercise 

(1) Prove that 

n 3 + Sn 2 + 2 . 
2n3 + 9 -in*--. 

(2) Decide for what values of x the limit 
[x+x n \ 



lim 



1 + x n 



exists. Draw a graph which plots the value of the limit against the value 
of*. 
(3) Use the sandwich theorem to prove that 

y/(n + 1) — y/n -> as n -*■ °°. 



Convergent sequences 



37 



" (4) Let x be a positive number, and let N be the smallest natural number 
satisfying N>x. Prove that 



^< 



x"- 



n\ (N-l)l \N 



n-N+l 



(n > N). 



Deduce that x n /n ! -*■ as n -* °°. 
' (5) Let a be any positive rational number and let \x\ < 1 . Show that there 
exists a natural number N such that (1 + 1/A0 a+1 |»l < 1. Deduce that 



1 x"\<\N a * 1 x 



a + l„iV| 



(n>N). 



Hence show that n a x n ■+ as n -*• °°. 
'(6) Prove that the sequence (n Vn ) decreases for B>3. [Hint: example 
4.19.] Hence prove that the sequence converges. 

(The convergence of <n 1/n > can also be established by a method like 
that of example 4.12. If n = (1 + /*„)", then it follows from the binom- 
ial theorem that 

n -^Hl<n. 

Therefore h n •* as « -* °°.) 



4.21 Some simple properties of convergent sequences 

4.22 Theorem A sequence can have at most one limit. 

Proof Suppose that x n -* I as n -» °° and x n -> m as « -> °°. Let e > be 
given. Then 

|f— »| = \l-x n +x n -m\<\l-x n \ + \x n -m\<e + e = 2e 

(triangle inequality) 

provided that n is sufficiently large. But, from example 1 .7 it follows that, if 
< ||/ - m\< e for every e > 0, then |/ - m\ = 0, i.e. l = m. 



4.23 Theorem Suppose that x„ ■+ / as n ■* °°. 
(i) If x„ > a (n = 1 , 2, . . .), then / > a. 
(ii) If x n <b(n = \,2,.. ), then / < b. 

Proof Kx n <b(n= 1,2,. . .), then -x„ >-b (n = 1, 2, 
we need only prove (i). 

Let e > 0. Then there exists an N such that, for any n>N, 

\x„-l\<e 

i.e. l-e<x n <l+e. 



.). Hence 



38 Convergent sequences 

But x n > a (n = 1 , 2, . . .) and so, for any n > N, a < x n < I + e. 

Hence, given any e > 0, a < / + e. From example 1 .7 it follows that a <. I. 

4.24 Example Suppose that x n ■* I as n -* °°. If x n > a (« = 1 , 2, . . .), it is 
tempting to conclude that l> a. But this may not be true. For example, l/n -*■ 
as n -*■ °° and l/n > (n = 1 , 2, . . .). But we certainly cannot conclude that 
0>0. 



4.25 Tiieorem Any convergent sequence is bounded. 

Proof Let x n -*■ I as n -*■ <*>. We have to find a ^ such that |jc„| < K 
(n = 1, 2, . . .). (See proposition 2.3.) 

It is true that, for any e > 0, there exists an N such that, for any n>N, 
\x„ — 1\ < e. In particular, this is true when e = 1, i.e. there exists an N t such 
that, for any n> JVj, 

Ix„-/I<l. 

From theorem 1 .18 it follows that, for any « > A 7 !, 

|jc b I-|/|<I*„-/|<i 

i.e. M<|l| + ] (n>iV,). 

The result now follows if we take 

K = max{|.r,|, \x 2 \, ..., |jf»|, |/| + 1}. 



4.26 Divergent sequences 

A divergent sequence is a sequence which does not converge. Any un- 
bounded sequence is therefore divergent (theorem 4.25). However bounded 
divergent sequences exist as well. 



4.27 Example The sequence <(— l) n > diverges. 

Proof The sequence is obviously bounded. Suppose that (— 1)" ■* / as 
n -*■ °°. Then, given any e > 0, we can find an N such that, for any n> N, 
l(~ 1 )" — /| < e. But there are both even and odd values of n greater than N and 
so 1 1 — /|< e and |— 1 — /| < e. These inequalities must be true for any e > and 
so we obtain the contradiction / = 1 and / = — 1 . 



We say that a sequence (x n ) diverges to + °° and write x n -*■ + °° as n-* °° if, 
for any H > 0, we can find an IV such that, for any n>N, 



Convergent sequences 



39 



x n >H. 

Similarly, a sequence <x n ) diverges to—°° and x n •* — °° as n -*■ °° if, for any 
H > 0, we can find an iV such that, for any n > N, 

x n <-H 



II - 




N 

- :i< r. - -- 




x„— *— «asn— * 



A divergent sequence <A" n > which does not satisfy x n -> + °° as n ■* °° or 
*,,->• — ooas«->°°is said to oscillate. An example is the sequence <(— 1)">. 



4.28 Example Let a be a positive rational number. Prove that if -* + °° as 
n -*°°. 



n>N, 



Proof Let H > be given. We have to find an A' such that, for any 



n a >H 

i.e. n>H ila . 

We simply choose A' = H Ua . Then, for any n > N, n >H va and hence n a > H. 



A common error is to suppose that proposition 4.8 (about combining con- 
vergent sequences) applies to sequences which diverge to + °° or — °°. For 
example, it is tempting to suppose that, if x n -* °° as n •* °° and y n •* °° as n •* °°, 
then x n —y n -»°° — °° = 0asn->°°. But °° is not a real number and cannot be 
treated as such. Indeed, one should immediately be warned that something is 
wrong by the appearance of the totally meaningless expression °° — °°. To press 
the point home, we consider the example x n = n 2 (n = 1, 2, . . .) andy„ = n 
(n = 1,2,...). Then x n -*• °° as n •* °° and y n -*■ °° as n -> °°, but x n - .y n "* °° as 
« ->°°. Or, again, takex„ = (n + 1) (n = 1, 2, . . .) andy n = n (n = 1, 2, . . .). 
This time jr n — y n -*■ 1 as « -* °°. 



40 



Convergent sequences 



4.29 Exercise 

(1) Suppose that x n -*■ I as n -> °°. Prove that 
(i) \x„ — 1\ -> as n -*■ °° 

(ii) \x n \ ■* |/| as« -*«>. 

(2) Suppose that x n -*■ / as « -+ °°. If / > 0, prove that there exists an N such 
that, for any n >N, 

x n >\l. 

(3) Prove that 

(i) 2" -*■ + °° as n -* °° (ii) — V" ""* ~~ °° as " "* °° 

(iii)<(-l)"n> oscillates. 

(4) Let jc„ >0 (« = 1 , 2, . . .). Prove that x„ -» as « ■* °° if and only if 
l/x n -*-°°as/! ^°°. 

(5) Suppose that {x n ) increases and is unbounded above. Prove that 

x n -*■ + °° as n ■* °°. If <x n > decreases and is unbounded below, prove 
that x n -> — °° as « * °°. 
* (6) Let 5 be a non-empty set of real numbers and suppose that d{%, S) = 
(see exercise 2.13(4)). Show that, for each n £ N, we can find an 
x n GS such that |£ — x n \ < l/«. Deduce that x n -> i- as « -> °°. 

If S is bounded above, show that a sequence of points of S can be 
found which converges to its supremum. If S is unbounded above, show 
that a sequence of points of S can be found which diverges to + °°. 



SUBSEQUENCES 



5.1 Subsequences 

Suppose that < x n ) is a sequence and that < n r > is a strictly increasing 
sequence of natural numbers. Then the sequence 

is called a subsequence of < x n ). 

The first few terms of some subsequences of < x n > are listed below. 

\X r+i ) = ^2,^3,^4, . . . 

'*3r ' = x 2> x 4> x 6> ■ ■ ■ 
'*3r+S' = x S> x ll> x 14i x n> ■ ■ ■ 
(X^r) = X2, Xn, -Xg, Xi6, . . . 

Roughly speaking, if we think of a sequence as a list of its terms, then we 
obtain a subsequence by crossing out some of the terms. For example, < x 2n ) 
may be obtained from <x n >by crossing out terms as below. 

Al> x 2> A3. • JC 4>/S' *6> ■ • ■ 

The insistence that < n r > is strictly increasing is of some importance. It means, 
for example, that the sequence whose first few terms are 

-*3> X \, X S, X \> X <)> ■ • • 

is not a subsequence of < x n ). 

The fact that (n r ) is strictly increasing has the consequence that 

n r >r (r = 1,2, .. .). 

This is easily proved by induction. 

5.2 Theorem Suppose that x n -»/ as n-*°° and that <x„ r > is a subsequence 
of (x n ). Then 

x n -* I as r •* °°. 

Proof Let e > be given. Since jc„ ■* I as n -* <*>, there exists an iVsuch 
that, for any n >N, 

41 



42 



Subsequences 



I *,-/!<«. 

Lei R=N. Then, for any r>R,n r >r> R = N and so w r > N. Thus 

We have shown that, given any e > 0, we can find an R (namely R = N) such 
that, for any r>R, 

\x„ F -l\<e. 

Hence x„ ■* I as /■ ■* °°. 



5. J Example We have seen that the bounded sequence ((— 1)"> diverges. 
A simpler proof can be based on theorem 5.2. Obviously the subsequence of odd 
terms tends to — 1 and the subsequence of even terms to + 1. But, if <(— 1)" > 
converged, all its subsequences would tend to the same limit. 

5.4 Example In example 4.4 we showed that, if x > 0, then x vn •* 1 as 
n -*■ °°. We prove this again by a different method. As in example 4.4 we need 
only consider the case x>l. 

If x > 1 , then x lln > 1 (n = 1 , 2, . . .) and hence the sequence < x lln ) is 
bounded below by 1 . Moreover 



I/n; l/(n+l) _ (n+l-n)/n(n+l) _ 



"/* 



= X 



= x 1/n(n+1) >l (« = 1,2,...) 



and therefore (x' ,n ) decreases. From theorem 4.17 it follows that <*""> con- 
verges to a limit / and / > 1 . What is the value of /? 

By theorem 5.2, we know that all subsequences of (x 11 ") must also tend to /. 
Hence 



,1/2" 



-*• / as n -*■ °°. 



Using proposition 4.8, it follows that 

But a sequence can only have one limit. Thus 

/ = I 2 

and so / = or / = 1 . But / > 1 and hence / = 1 . 
We have shown that 



Aln 



-*■ 1 as n -*■< 



Note: A warning is appropriate here. Do not use this method of working out the 
value of a limit until you have shown that the sequence converges. For example, 
if you thoughtlessly tried to evaluate the limit of the sequence <(-- 1)"> by 
looking at its subsequences, you would obtain / = + 1 = — 1. 



' 






Subsequences 



43 



5.5 Example Let a >0 and let jc, >0. Define the rest of the sequence <x n ) 

inductively by 

*„ +1 = Kx n + OX?) in = 1,2,...). 
Obviously x n >0(n = 1,2, . . .). But further 

x 2 n -2x Ml x n +a = 0. 

We know in advance that this quadratic equation in x n has a real solution. 
Thus 'b 2 - 4ac > 0' , i.e. 

x 2 +l >a. 

Since jc n+1 > 0, it follows that 

x n+l >s/a (n = 1,2,...). 

We wish to prove that < x„ > decreases and so we consider 

* n -*n+i = x„-i(x„+ ax- 1 ) 

> (n = 2,3,...). 

Hence the sequence <jc„> decreases and is bounded below by Va (provided 
we omit its first term). From theorem 4.1 7 it follows that 

x n -* I as n -* °° 

where / > \Ja. What is the value of/? 

By theorem 5.2 we also have x n+1 -* / as n -* °°. But 

Now / ^ (because / > \Jd). Hence, by proposition 4.8, 

x n +i = i(* n + ^n 1 )- > K' + «'" , )asn^ 00 - 
Since a sequence can have at most one limit, 

/ = kQ + al" 1 ) 

i.e. I 2 = a. 

Thus / = yJaoxl = — \Ja. But we know that / > \/a. Hence / = \/a. We have 
therefore shown that 

x n -* \Ja as n -* °°. 



44 



Subsequences 



5.6 Example We attack the previous example by a different method. Again 

let a > but this time let Xi be unspecified for the moment. Define the rest of 
the sequence inductively by 



Now 



x n +i = ttxn+ax- 1 ) (n = 1,2,...). 
x n+l ~\/a = \(x n + ax~ l )-s/a 



2x, 



(x 2 n -2x n y/a + a) 



' 2x n \ 2jc n _, 



1 



1 
2x n (2* n _,) 2 



{x M - Va) 4 
1 



1 



1 



1 



2x n (2x n „ 1 ) 2 ---(2x i y 



P5<*i-v«r. 



If we assume that Xj > \/a, then it follows, as in example 5.5, that x n > y/a 
(n= 1,2, . . .). Hence 



l* n +i-\/«l < 



2%Ja 



2y/a 



l + 2+2 2 +...+2" _1 



c^-v«r 



(2"-l)/(2-l) 



(*i-vfc)* n 



= 2Va 



■ i~ Va 

2Va 



We know that, if 1 7 | < 1 , theny" -> as « ■+ °°. By theorem 5.2 it follows 
thaty 2 -*0asn-*°°. 

Hence our argument shows that x n •* \Jaa.sn^>°° provided that 



2\/« 



<1. 



We have already assumed that x x > \Ja and so we have shown that x n -* V<z as 
« -* °° provided that \]a <x y < 3 \Ja. Of course we already know from example 
5.5 that this was true under the weaker hypothesis that x x > 0. However, the 









Subsequences 



45 



argument given here is interesting because it allows us to estimate how 'good' an 
approximation x n is to \/a. 

For example, suppose we want to estimate the value of y/2. It is obvious that 

1 < sj2 < 2 (because l 2 = 1< 2 and 2 2 = 4 > 2). For a better estimate take a = 

2 and*! = 2 in the argument above. Then 



|x 4 -V2|«2V2 
2V2 



(2-V2 



2V2 



= 2\/2 



V2(V2 - 1) 



2V2 



(V2-1) 8 . 



Using the crude estimate \J2< 2 we obtain |x 4 — \/2 |<2 _6 = p< 0-016. 
Hence x 4 = gg = 1 414 . . . differs from y/2 by at most 0016. 

(A better estimate for \J2 may be obtained by evaluating x s or by starting 
with Xj = \ and using the estimate \J2 < §.) 



5. 7 Exercise 

(1) Given that < n u " > converges (see exercise 4.20(6)), show that 

/i l/n -> 1 asH-* 00 

by considering the subsequence ((2m) 1 ' 2 "}. 

(2) A sequence < x n > is defined by x x = h and 

x n*l = x n ' " 

where < k < \ and h lies between the roots a and b of the equation 
x 2_ x+k=Q 

Prove that a <x m . 1 <X n < b {n = 1, 2, . . .). Show that (x n ) con- 
verges and determine its limit. 
* (3) If k > and x x > and ( x n > is defined inductively by 



■*n+1 — 



l+x n ' 



show that one of the sequences <,x 2n > and (x 2n -i > is increasing and 
the other decreasing. 

Prove that both sequences converge to the limit / which is the posi- 
tive root of the equation x 2 + x = A:. What conclusion may be drawn 
concerning <*„>? 
' (4) Two sequences < x n > and (y„ > are defined inductively by x l = \ and 
y-i = 1 and 

X„ = VCXn-lJ'n-l) (» = 2,3,...) 



46 



Subsequences 






2 W„ 



J. 

y n -i) 



(n o 2,3,...)- 



Prove that x„_, <x„ <^ n <>>„_, (« = 2, 3, . . .) and deduce that both 
sequences converge to the same limit /, where ? < / < 1 . (Actually / = 
ff/4.) 
1 (5) Let y be any real number. Prove that the sequence 

<(l+yn- l ) n ) 

is increasing for those values of n which satisfy n > 1 —y. [Hint: 
example 4.19.] Show also that the sequence is bounded above and 
hence converges. [Hint: From exercise 4.20(4), (2y)"/n\ ->0 as n -> °° 
and therefore we can find an N such that, for any n>N,y"/nl< (£)".] 
' (6) Show that the product of the limits of the two sequences <(1 + xn' 1 )") 
and <(1 -xn'')") is equal to 1. 



'" 1 + « V~7i 



i-$ 



n 2 \l/n-i 



5.8 



5.9 



Bolzano-Weierstrass theorem 



Theorem Every sequence has a monotone subsequence. 
Proof Let (x n > be any sequence of real numbers. We must construct a 
subsequence {x„ r ) which is either increasing or decreasing. We distinguish two 
cases. 

(i) Every set {x n : n > N} has a maximum. In this case we can find a sequence 
(n r ) of natural numbers such that 

x n = maxjc n 

n>l 

*„,, = maxx n 

n>n, 

x„ 3 = max x n 

and so on. Obviously n t <n 2 <n 3 < . . . and so, at each stage, we are taking 
the maximum of a smaller set than at the previous stage. Hence < x n > is a 
decreasing subsequence of < x n > . 

(ii) Suppose that it is not true that all of the sets {x n : n > N} have a maxi- 
mum. Then, for some jV, , the set {x n :n>N t } has no maximum. It follows 
that, given any x m with m>N lt w can find an x n following x m such that 
x n >x m . (Otherwise the biggest of x N +1 , . . . ,x m would be a maximum for 
{*„:«>AM). 



47 






Subsequences 

We define x ni = x Ni+1 and then let x„ 2 be the first term following *„, for 
which x n >x n '. Now let x nj be the first term following x„ 2 for which 
x n >x n . And so on. We obtain an increasing subsequence <*„ r > of (x n ). 

5.10 Theorem {Bolzano-Weierstrass theorem). Every bounded sequence of 
real numbers has a convergent subsequence. 

Proof Let < x n > be a bounded sequence. By theorem 5.9, (x n > has a 
monotone subsequence <*„,.>. Since (x n ) is bounded, so is <x„ r >. Hence, by 
theorem 4.17, <x„ r > converges. 



5.11 Examples 

(i) The sequence <(— 1)" > is bounded. Two convergent subsequences are the 
sequence — 1 , — 1 , — 1 , ... of odd terms and the sequence 1 , 1 , 1 , 1 , . . . of 
even terms. 

(ii) A more sophisticated example is the sequence ( r n > whose first few terms 

are . 

2' 3> 1' 4> 4i $> S< ?i 5> 5> Sj • ■ • 

Every term of the sequence is a rational number between and 1 . Hence the 
sequence is bounded. It has many convergent subsequences. For example 

5, 4> 6> S. • • • 

ill! 

2> 3. 4> Sj ■ • ■ 

uu... 



5.12* Lim sup and lim inf 

We shall not have a great deal of use for the subject matter of this 
section. However, it is material which has to be learned eventually and it is 
natural to deal with it here. 

Suppose that <jc„> is a bounded sequence and let L denote the set of all real 
numbers which are the limit of some subsequence of < x n >. We know from the 
Bolzano-Weierstrass theorem that L is not empty (i.e. L # 0). Of course, if 
(x„) converges, L will just consist of a single point (theorem 5.2). 

5.13^ Proposition Let < x n > be a bounded sequence and let L be the set of 
all real numbers which are the limit of some subsequence of < x n >. Then L has a 
maximum and a minimum. 

That L is a bounded set follows from theorem 4.23. Hence L certainly has a 
supremum and infimum. Proposition 5.13 asserts that these actually belong to 
the set L. The proof is relegated to the appendix. Readers who prefer to omit 



48 



Subsequences 



the proof altogether will encounter it again at a later stage since it is a special 
case of a general theorem concerning 'closed sets'. 

If (x n > is a bounded sequence, let us denote the maximum of L by 7 and the 
minimum by /. Then / is the largest number to which a subsequence of <x„> 
converges and / is the smallest number to which a subsequence of < x n > con- 
verges. 

We call / the limit superior of ( x n > and write 

/ = lim sup x n . 

n—*oo 

Similarly, / is the limit inferior of <x„ > and we write 
/ = lim inf x„. 

(Occasionally you will see such expressions as lim sup x n = + <*>. This simply 
means that < x n > is unbounded above. Similarly lim inf x n = - °° means that 
(x n ) is unbounded below.) 



5.14 * Examples 

(i) The sequence < 1/n > converges with limit 0. By theorem 5.2, all its sub- 
sequences converge to 0. Hence L contains the single point and 



lim sup — = lim inf — = lim — = 0. 



(ii) The sequence <(— 1)"> is bounded. The set L consists of two points + 1 
and-1 (i.e.Z, = { + 1, - 1}). We have 

lim sup (-1)" e 1; lim inf (-1)" = -1. 



M,U, 



(hi) Let < r n > be the sequence whose first few terms are \, J, §, „, , 
I, \, . . . Obviously <r n < 1 (n = 1, 2, . . .). From theorem 4.23 it follows that 
L is a subset of [0, 1 ] (i.e. L C [0, 1 J). But the subsequence \, 1,1,1,... con- 
verges to 0. Thus / = 0. Also the subsequence £, \, |, f, . . . converges to 1 . Thus 
/ = 1 . We have shown that 

lim sup r n = 1 ; lim inf r n = 



5.15* Exercise 
*(!) Calculate 



(i)lim sup {(-!)"(!+ «-»)} (ii) lim inf {(-1)"(1 + «"')} 



Subsequences 

and show that these are not the same as the numbers 



49 



sup (-1 )"(! + « _1 ) and 



inf (- 1)"(1 



»-»). 



* (2) Let < r n > be the sequence whose first few terms ate §, hhk, $»t*h 5* I* 

|,4» .. . Show that every point in the interval [0, 1] is the limit of a 
subsequence of <r„>. [Hint: see exercise 3.6(4).] 
t (3) Suppose that (x n ) is a bounded sequence and that, for any N, we can 
find an n > N such that x n > b. Show that < x n > has a subsequence 
which converges to a limit / > b. 

* (4) Let < x„ ) be a bounded sequence with limit superior / and limit inferior 

/. Show that, given any e > 0, we can find an N such that, for any 
n>N,x n <l + e. [Hint : use question 3 with b = / + e] . What is the 
corresponding result for /? 
t (5) Deduce from question 4 that a bounded sequence < x n ) converges with 
limit / if and only if 

lim sup x n = lim inf x n = I. 

Hence show that a bounded sequence < x n > converges if and only if all 
its convergent subsequences have the same limit. 

* (6) Let < x„ > be a bounded sequence and let 

,M n = sup x k . 

Show that < M n ) decreases and is bounded below. Deduce that (M n ) 
converges and denote its limit by M. If / is the limit of some subse- _ 
quence of < x n >, show that M n > I (n = 1 , 2, . . .). Deduce that M > I. 
Obtain also the reverse inequality M < I [Hint: use question 4] and 
hence show that 

lim sup*,, = lim {sup x k }. 

(This explains the choice of notation for the limit superior. We also 
have, of course, 

lim inf x n = lim { inf x k } .) 



5.16 Cauchy sequences 

Suppose that we want to prove that a sequence < x n > converges but we 
have no idea in advance what its limit / might be. Then there is no point in 
appealing directly to the definition because we shall certainly not be able to 
prove that \x„ — I \ can be made as small as we choose by taking n sufficiently 
large if we do not know the value of /. 



50 Subsequences 



Theorem 4.17 gives us one way of resolving this difficulty. It tells us that an 
increasing sequence which is bounded above converges and that a decreasing 
sequence which is bounded below converges. Thus we are able to deduce the 
convergence of the sequence without necessarily knowing the value of its limit. 
But, of course, theorem 4.17 deals only with monotone sequences. In order to 
deal with sequences which may not be monotone, we introduce the idea of a 
Cauchy sequence. 

We say that < x n ) is a Cauchy sequence if, given any e > 0, we can find an N 
such that, for any m >iVand any n>N, 

\x m —X n \<e. 

Very roughly, the terms of a Cauchy sequence get 'closer and closer' together. 

5.17 Proposition Any convergent sequence is a Cauchy sequence. 

5.18 Proposition Any Cauchy sequence is bounded. 

The proofs are easy. The next theorem is what makes Cauchy sequences 
important. 

5.79 Theorem Every Cauchy sequence converges. 

Proof Let (x n ) be a Cauchy sequence. By proposition 5.18, (x n > is 
bounded. Hence, by the Bolzano-Weierstrass theorem, it has a convergent sub- 
sequence (x nr >. Suppose that x„ r -*■ / as r -*■ °°. We shall show that x n -* 1 as 

Let e > 0. Then \e > 0. Hence there exists an R such that, for any r >R, 



„ r -/K£e. 



0) 



Since < x n > is a Cauchy sequence, there also exists an N such that, for any m>N 
and any n>N, 

\Xm-X n \<y. (2) 

Now suppose that n > N and choose r so large that n r >Nandr>R. Then 
(1) is satisfied and also (2) is satisfied with m = n r . Thus, for any n > N, 

\x n -l\ = l**-*„ r + ;c nr -/| 

< I x n — x„ r | + \x„ r — / 1 (triangle inequality) 

< ^e + ie= e. 

Given any e > 0, we have found an A'' such that, for any n > N, \x„ — 1 1 < e. 
Thus x n -> / as n -> °°. 



Subsequences 

5.20 Example A sequence <x n ) is defined by x, = a, x 2 = b and 
x n + 2 = |(Xn+l + *n) (" = 1,2,3,...). 

Prove that <x n > converges. 
Proof We have 

X n *2~ x n+\ = iOn+l +X n) ~ x n-H = h( x n ~ x n+l)- 



51 



Thus 



\X„+2~ x n*l I a «l*« x n+\\ 2 |A " "-' 



— ^,2 ' X " X "-l 



1 1 

— \x 2 -x l \ = 



2 n 



Hence, if n >m, 

\Xm—X»\ = \x n -x n ^+x n _,-x„_ 2 -.. .+x m+i -x ri 



*tl ••mi 



< I X n -X„_ t I + I X n _, -X n -2 i + • • ■ + I x m+l ~ x n 



i=i+=i+.-- + 5^: 



2 m - 1 1 - 1 
Let e > be given. Choose N so large that 
1 



'4 + ? + - + 5^' 



1 rt",-,Kil»-,l. 



,N-2 



Lft —«l<«. 



Then, for any n > N and any m > N. 

Thus < x n > is a Cauchy sequence. Therefore, by theorem 5.19, it converges. 

(An alternative method would be to show that one of the two sequences 
<*2n-i> and <*2*> is increasing and the other decreasing and then to show that 
they have the same limit.) 



5.21 Exercise 

(1 ) Suppose that < a < 1 and that < x n > is a sequence which satisfies 
I x n+i ~ x n I < a " (" ~ 1 ■ 2 - • • •)• Rmw that < x " ) is a Cauch y sequence 
and hence converges. 



52 



Subsequences 

Give an example of a sequence < y„ > such that y n ■* + oo as n _► oe 
but lj' n+1 ->' n |-*0asM-*oo. [Hint: see exercise 4.20]. 
(2) A sequence <x„> satisfies 0<a <x, <x 2 < b and 

* n+2 = {*n + i*„}" 2 (« = 1,2,...). 

Prove that a <x n < b (it = 1 , 2, . . .). Hence or otherwise show that 

6 



I 



I ***! -*»!<■ 



-*n -**3 



(« = 2,3,...). 



Deduce that <*„> is a Cauchy sequence and hence converges. 

(3) For the sequence of example 5.20, prove that 

x n+i + hx n = x n + \x n _ i = ... = x 2 + \x { . 

Deduce that | x n+i -l\ = l\x n -l\, where / = |(jc 2 + £*i> What con- 
clusion may be drawn about the convergence of the sequence <*„>? 
Tackle exercise 5.21(2) by a similar method. 

(4) Let [a, b] be a compact interval (see §2.9). Prove that every sequence 
of points of [a, b] contains a subsequence which converges to a Doint 
of [a, b]. 

» (5) Let /be an interval which has the property that every sequence of 
points of /contains a subsequence which converges to a point of/. 
Prove that /is compact. [Hint: First prove that /is bounded by assum- 
ing otherwise and appealing to exercise 4.29(6). Then prove that sup / 
and inf /are both elements of /with another appeal to exercise 
4.29(6).] 

T (6) Given a set S of real numbers, let 

5j = {x:xGSandx=t$. 

We say that | is a cluster point (or point of accumulation or 'limit 
point') of £ if £ is at zero distance from S±. (Note that % need not be an 
element of S). A form of the Bolzano- Weierstrass theorem asserts that 
every bounded set with an infinite number of elements has at least one 
cluster point. Prove this. 



SERIES 



6.1 Definitions 

Given a sequence <c n > of real numbers the sequence <s N ) defined by 



hf ~ L a n ~ fl l + fl 2 + • ■ • + "N 

is called the sequence of partial sums of the series 
Z a n . 

n = l 

If s N ■* s as N -» °°, the series is said to converge to the sum s . We write 
S = £ a n . 

n = l 

It is important to remember that this formula can only make sense when the 
series converges. 



6.2 Example Consider the series 



n = o 

If x J= 1 , the partial sums satisfy 



Sfl = £ x n = 1 + x + x 2 + . . . + x 

n = 



N 



1 -x 



.\'+ 1 



1 -x 

If |x|< l,.r N+1 -*0 as A'-* °° and hence 



53 



54 



Series 



Series 



55 



s N - 



1 



as N -*■ °°. 



1 -* 

It follows that the series 2~ =0 x" converges if |jc| < 1 and we may write 

1 



!>" = 



l-x 



(W<1). 



If \x I > 1 , the series diverges. 



6.3 Example The partial sums of the series 

CO J 

I — — 

n = i n(n+ 1) 
are given by 






-l|i- 



n = 1 n(n + 1) n = i \n n 4- 1 



4WBWH+ 



l 



i 



N N+ 1 



= 1 -• 



1 



--> 1 asAf-* 00 . 
N+ 1 

Hence the series converges and we may write 



1 



n = \n(n + 1) 



= 1. 



6.4 Series of positive terms 

Series whose terms are all positive (or non-negative) are particularly 
easy to deal with. This is because the sequence of partial sums of such a series is 
increasing. Thus, if we wish to show that a series of positive terms converges, we 
only need to show that its sequence of partial sums is bounded above. If the 
sequence of partial sums is unbounded above, then the series diverges to + °° 
(see exercise 4.29(5)). 

6.5 Theorem The series 

n = i n 
diverges to + °°. 



Proof Since the partial sums of this series increase, we only need to 
show that they are unbounded above. But 



w 



s 2N = I+^ + t+.-. + ^v 






i 

2 N 



4 4 



iN-l 



Thus the partial sums are unbounded above and the theorem follows. 
6.6 Theorem Let a be a rational number such that a > 1 . Then the series 

GO 1 

converges. 

Proof Since the partial sums are increasing, we need only show that 
they are bounded above. For TV > 1, 



1 . 1 



1 









+ 



i 



T(JV-l)a 



+ . . .+ 



1 



(2 N -lfj 



< 1 



;<* 2° \4 a 4" 4" 4 a ! 



+ 



1 



2<fi-i)u 



+ ...+ 



1 



2(N-l)Of 
9JV-I 



2 4 



= 1 + 



2 a 
1 



+ 



1 



2 «-l ■ ^or- 
al - 1\JV 



+ . . .+ 



N-l 



l-(l/2 a - 1 )^l-(l/2 a - 1 ) 



56 Series 

provided that l/2 a_1 < 1 . But this follows from the assumption that a > 1 . 
Hence <s^> is bounded above and the theorem follows. 



6.7 Elementary properties of series 

6.8 Theorem Suppose that the series 2„ = i a„ and S" =1 b n converge to a 
and respectively. Then, if X and p. are any real numbers, the series 

2„ = i (ka n + pb n ) converges to A a + ju0. 

Proof We have 
IMA) = X la n +n £>„ 

n = l ,. = 1 n = l 



-+\a + p@ asiV-x* 5 



by proposition 4.8. 



6.9 Theorem Suppose that the series 2„ = i a n converges. Then 

a n -> as « ■* oo. 
Proof Let the sum of the series be x. Then 

N 

s n = Z a„-»-sas?j->°°. 

n = l 

Also x.v _ l -*■ s as w -* <*>. 
But then 

flw = (fli +a 2 + • • • + %) - (fli + a 2 + • • • + %-j) 
= s N — s N - l -*s— s = OasTV-*- 00 . 

6.70 Examples The series 

X(-D" 

n = l 

diverges. This can be deduced from theorem 6.9 by observing that its terms do 
not tend to zero. 

Note that the converse of theorem 6.9 is false. Just because the terms of a 
series tend to zero it does not follow that the series converges. Theorem 6.5 
provides an example. The terms of the series 

OS 1 

I 1 
n=i n 



Series 57 

tend to zero, but the series diverges. One might say that the terms of the series 
do not tend to zero 'fast enough' to make the series converge. 



6.11 Proposition Suppose that the series S" =1 a„ converges. Then, for each 
natural number TV, the series £„ = /v a n converges and 

CO 

Y, fl n -*0as7v'-»-°°. 

n=N 

This result is often referred to by saying that the 'tail' of a convergent series 
tends to zero. The proof is easy. 



6.12 Series and Cauchy sequences 

In the series we have studied so far, we have either been blessed with a 
nice formula for the partial sums (examples 6.2 and 6.3) or else the partial sums 
increased and so we only had to consider whether or not the partial sums were 
bounded above (theorems 6.5 and 6.6). What should we do in the absence of 
such favourable conditions? We ask the question: is the sequence of partial sums 
a Cauchy sequence*! This question has some fruitful answers as we shall see be- 
low. 

6.13 Theorem Suppose that <a„> is a decreasing sequence of positive num- 
bers such that a„ -> as n -*■ °°. Then the series 

oo 



converges. 

Proof We show that the sequence <s„> of partial sums of the series is a 
Cauchy sequence. From theorem 5.19, it then follows that the series converges. 
The proof depends on the fact that, for each n>m,we have the inequality 

0<a m + 1 — A m + 2 +a m + 3 — . . .<2„<fl m + 1 . 

This follows easily from the fact thata*, — aj, + 1 is always non-negative because 
<«(,> decreases. 

Let e > be given. Since a n -*■ as n -* °° we can find an N such that, for any 
n>N,a n <e. But for any n>m>N, 

\$n~ *ml = lfai-«2 +a 3 -. . .a n )-(a,—a 2 + . . .a m )l 

= la m + i — a m+2 +a m+3 — . . .a„\ 

< fl m + i<e (because m >N). 

Thus <s„> is a Cauchy sequence and the theorem follows. 



58 



Series 



6.14 Example We saw in theorem 6.5 that the series £™ =1 l/« diverges. But 
it follows from theorem 6.13 that the series 

n •■ i n 2 3 4 5 

converges. Of course, theorem 6.13 yields no clue as to what the sum is. (It is, in 
fact, logg2.) 



Theorem 6.13 is sometimes useful, but its hypotheses are rather restrictive. A 
more useful theorem is the following. 

6.15 Theorem (comparison test) Let 2™ =1 b n be a convergent series of posi- 
tive real numbers. If 

\a n \<b„ (« = 1,2,...) 

then the series 2„ = i a n converges. 

Proof Let e > 0. Since 2~=i b„ converges, its tail tends to zero (prop- 
osition 6.1 1). Hence we can find an A' such that, for any n >N, 

I b k <e. 

ft=n + l 

Let the sequence of partial sums of the series Z" = , a n be <s„>. Then, if 
n>m>N, 

!»«— %l = \(ai + ai + . ..+a n )-(ai + a 2 + . . . + fl m )l 

< |a m + il + |fl m + 2 | + . . . 4- \a„\ (triangle inequality) 



< I *fe<e. 



Thus (s„) is a Cauchy sequence and the theorem follows. 

Note that the hypothesis \a„\ < b n (n = 1 , 2, . . .) can be replaced by 
\a n \^Hb„ (n =N,N+ 1, . . .). (Why?) Note further that the hypothesis of the 
comparison test is not 



fe=i 



< I**. 



This condition is satisfied by a k = (— 1)'' and & ft = l/k 2 , but we know that 
2" =1 (- 1)" diverges. 



Series 59 

6.16 Example Prove that, if a is any positive rational number and \x\ < 1, 
then the series 

n = l 

converges. 

ft-oo/ From exercise 4.20(5) we know that n a+2 x" -+0zsn-*°°. 
Hence the sequence (n a+2 x n ) is bounded (theorem 4.25). It follows that, for 
some H, 

\nV\<H/n 2 (m = 1,2, . . .). 

Since the series S~ =1 l/« 2 converges (theorem 6.6), the result follows from 
the comparison test. 



The next two propositions are sometimes helpful. We indicate only very brief- 
ly how they are proved. 

6.17 Proposition (ratio test) Let S„ = x a„ be a series which satisfies 
fln + i 



lim 



a* 



= I. 



If / > 1 , the series diverges and, if / < 1 , the series converges. 

If /< 1, we may take e>0 so small that /+ e< l.Then, for a sufficiently 
large value otN, 



\a„\ = 



a„ 


"n-2 




°jV + 2 


"n-l 



\a N -n\<Q + ey- N - l \a N ^\. 



The series S" =1 a n then converges by comparison with the geometric series 
2" = i (/ + e) n . If / > 1 , a similar argument shows that the terms of 2™ = i a n do 
not tend to zero and so the series diverges. 

6.18 Proposition (nth root test) Let ~L" =i a n be a series which satisfies 
lim sup \a n \ i,n m I. 

If / > 1 , the series diverges and, if / < 1, the series converges. 

If / < 1 , we may take e > so small that I + e < 1 . Then, for a sufficiently 
large value of TV, 

\a n \ <(l+ ef (n >N) (exercise 5.15(4)) 

and the convergence of 2"=i a n follows from the comparison test. If / > 1, e is 



60 



Series 



chosen so that / — e > 1 . Then, for some subsequence (a nk \ 

' a "fe' > Q ~ e )" fe "* °° as k "* °° 
and so the terms of S" =1 a„ do not tend to zero. 

Note that if the expressions of propositions 6.17 and 6.18 diverge to + °° 
(instead of converging to /), then the series diverge. If / = 1 however the results 
yield no information about the convergence or divergence of the series at all. 

6.19 Example Prove that the series 



T - 



converges for all values of x. 

Proof We could use the comparison test. Alternatively, if* ¥= 0, 

■ as n -*■ °° 



(n + 1)! / n\ 



n + 1 



and hence the series converges by the ratio test. 



6.20 Absolute and conditional convergence 

A series 2~ =1 a„ is said to converge absolutely if the series 2"=i \a n \ 
converges. A series which converges but does not converge absolutely is said to 
be conditionally convergent. 

6.21 Theorem Every absolutely convergent series is convergent. 
Proof Simply take b n = \a n \ in the comparison test. 



6.22 Examples 



(i) Let a n = '• — (n =1,2,...). Then 



„-i n =, n 2 3 4 



S ^1 111 



Series 



61 



The first of these series converges (example 6.14). The second diverges (theorem 
6.5). We conclude that the series 



(-1) 



n-l 



is conditionally convergent. 

(ii) On the other hand the series 



I*n=I 



(-1)" 



n = l n = l fl 

is absolutely convergent because 

I M = I k 

n=\ n=l n 

which converges. 



It should be noted that the comparison test, the ratio test and the nth root 
test all demonstrate absolute convergence. The only criterion we have given 
which can establish the convergence of a series which is only conditionally con- 
vergent is theorem 6.13 



6.23 Manipulations with series 

Series are 'infinite sums'. It would therefore be optimistic to expect to 
be able to manipulate them just like 'finite sums'. Indeed, only absolutely con- 
vergent series may be freely manipulated. If one tries to obtain results by 
manipulating divergent or conditionally convergent series, only nonsense can be 
expected in general. 



6.24 Example Consider the following argument. 
= + + + .. . 

= (1-1) + (1-1) + (1-1)+... 

= 1-1 + 1-1 + 1-1 + .. . 

= 1 +(-1 + l) + (- 1 + l) + (-l + 1) + 

= 1 +0 + + 0+ ... 

= 1. 



62 



Series 



The error is not hard to find. The series 1—1 + 1 — 1 + 1—1 
divergent. (Its terms do not tend to zero.) 



... is 



6.25 Example We know that the series 

1 2^3 4^5 6^7 •■• 

converges conditionally. Denote its sum by s. We rearrange the order in which 
the terms of the series appear and obtain 

i _1_i4.1_1_i4.1_1_i4.l_ 

1 2 4^3 6 8^5 T5 15^7 ••• 

Let the Mth partial sum of this series be t n . Then 

1 



1 1 1 



1 



1 



2m -1 



1 



In - 1 An - 2 

-|I + i+... 
2 6 



_1_ 

4m 



1 



4m -2 



1 1 

- - + -+. . . + — 
\4 8 An 

1 1 

1 +-+ . . . + 

3 2/1-1 



1 + 3 + 



+ 



2m -1 



1/11 1 

- + -+ . . . + — 

2 \2 4 In 



1 / 1 I 

= - 1 -- + +. .. 

2 \ 2 3 4 



2m - 1 2m 



\s as n 



Notice that the finite sum for t 3n can validly be rearranged in any way we like. 
Since t 3n + 1 — t 3n •* as n -*■ °° and r 3n+2 — r 3n ->0 aSM -*«»,it follows that 
the rearranged series converges to \s which is not the same as the sum s of our 
original series (because s ¥> 0). 



6.26 Exercise 

(1) Using partial fractions, prove that 

f 3n ~ 2 -1, 

M m(m + 1)(m + 2) 

(2) If <a„> and <b n ) are two sequences of positive terms and 



Series 



63 



► / as m -*■ °° 

b n 

where / <¥• 0, prove that the series 

£ a„ and £ b„ 

either both converge or both diverge. 

Discuss the convergence or divergence of the series 

(3) Suppose that <a„> is a decreasing sequence of positive terms such that 
_„ = ! a n converges. Prove that na„ -* as n -» °°. [ Hint: consider 

%*i+*»+a + • .. + «&»•] 

(4) Let </•„> denote the rational numbers from (0, 1) arranged in the 
sequence whose first few terms are \, 3, §, 4,4,4,... Prove that the 
series 



_/« 



diverges. 
(5) Determine whether or not the following series converge or diverge. 



®£_? 

n = i(2n)! 



ft Cm' - ) 2 



(iii) _ 



M + 



(iv) Z 

n = 



n = l \n + \j 
( _ ir -. 



4" (v) _. -{V(«+0-V«} 

n = l M 



n = l 



v« 



(6) If the sum of the conditionally convergent series 

I-i + 1-i + l-... 

iss, prove that the sum of the rearranged series 
1+1-1 + 1 + 1-J + J + X-1 + ... 



is Is. 



Hint:r 3n = 1+ J _ 2 + '" 



1 



+ 



4n — 3 4m - 1 2m 



7 



FUNCTIONS 



7.1 Notation 

A function f from a set A to a set fl (write /: A-*B) defines a rule 
which assigns to each x£ A a unique element y G.B. The element y is called the 
image of the element x and we writer = f(x). 

When A and B are sets of real numbers we can draw the graph of the function 
as in the diagram below. The defining property of a function ensures that each 
vertical line drawn through a point of A cuts the graph in one and only one 
place. 



/(-*) 




If / is a function from A to B and S C A , we say that / is defined on the set 
S. The largest set on which / is defined is, of course, the set A . We call A the 
domain of/. For example, a sequence is a function whose domain is the set M 
of natural numbers. 



/»f« 



» / \ 




4 — w&$ ) 



64 



Functions 



65 



If / is defined on a set S, we use the notation 
AS) = {f(x):xeS} 

and say that f(S) is the image of the set S under the function /. 

The set f(A) is called the range off. Note that f(A) need not be the whole 
offi. 



7.2 Example Consider the equation y = x 2 . This defines a function from 
U to itself. For each j£R there exists a unique y€U which satisfies the rule 
y — x 2 . Observe that, in the diagram below, each vertical line cuts the graph in 
one and only one place. 




The domain of this function is R. The range is [0, °°). The image, for example, 
of the set [-2, 1] is [0, 4]. (Why?) 



7.3 Example Consider the equation y 2 = x. This does not define a func- 
tion from R to itself. In the diagram below, the vertical line drawn through the 
point x does not meet the graph at all, and hence there is no value of y associ- 
ated with x . 




7.4 Example Again consider y 2 = x. Does this define a function from 
[0, °°) to R? Again the answer is no. This time it is certainly true that every 



66 



Functions 



vertical line drawn through a point of [0, °°) meets the graph (see § 1 .9). But all 
but one of these vertical lines meets the graph in two points. Thus, in the dia- 
gram, there is not a unique value of y associated with Xj. 




7.5 Example Again consider y 2 — x. This does define a function from 
[0,~)to[0,°°). 



10.°°) 




Since y must be in [0, °°) we omit from the diagram the part of our previous 
graphs which lies below the *-axis. Then every vertical line drawn through a 
point of [0, °°) meets the graph in one and only one point (see § 1 .9). Thus, 
given any x£ [0, °°), there is a unique y G [0, °°) which satisfies y 2 = x. Thus a 
function /is defined from [0, °°) to [0, °°). Recalling the content of § 1.9, we 
observe that, for each x > 0, 



7.6 Polynomial and rational functions 

If a ,a u a 2 , ■ ■ . a n are all real numbers, then the equation 

y = a + a t x + a 2 x 2 + . . . + a n x" 

defines a function from R to itself. Any value of x which is substituted on the 
right hand side generates a unique corresponding value of y. If a n # 0, we call 



Functions 



67 



this function a polynomial of degree n. A polynomial of degree is called a 

constant. 

Suppose that P and Q are polynomial functions. Let S denote the set R with 
all the values of x for which Q(x) = removed. (If Q is of degree m, it follows 
from exercise 3.1 1(3) that there can be at most m such values.) Then the equa- 
tion 

P(x) 



y = 



<m 



defines a function from S to U. Such a function is called a rational function. 



7. 7 Example The function from U to itself defined by the equation 

y =x 3 — 3x 2 + 2x is called a polynomial function of degree 3 (or, more loosely, 

a 'cubic polynomial')- Its graph is sketched below. 




= v'-ix 1 + 2.Y 



7.8 Example Let S be the set R with 2 and - 2 removed. Then the equa- 
tion 



y = 



x z +4 
x 2 -4 



(x*±2) 



defines a function from S to R. Its graph is sketched below. 




.v ! +4 



68 



Functions 



7.9 Combining functions 

We begin with some almost obvious notation. If S C IR and / and g are 
two functions from S to U , then we define the function /+ g to be that func- 
tion from S to R which satisfies 

(f+g)M = f(x)+g(x) (xes). 

Similarly, if X is any real number, we define X/ to be the function from S to IR 
which satisfies 

(x/)(x) = x/(x) (xes). 

Again, we define the functions fg and f/g by 
(.fg)(x) = f(x).g(x) (xes) 
(f/g){x) = f(x)lg(x) (xGS). 

For the latter definition to make sense, of course, it is essential thatg(x) =h 
for all x&S. 

A somewhat less trivial way of combining functions is to employ the oper- 
ation of composition. Let 5 and T be subsets of R and suppose that#: S-*T 
and /: T-* R . Then we define the composite Junction f°g:S->- R by 

/°*(*) = /(*(*)) (*es). 

Sometimes fa g is called a 'function of a function'. 



7.10 Example Let/: 



be defined by 



f<*)-$ + \ (* 


SR) 






and let g: R -*■ R be defined by 








*(*) = * 3 - 








Then fog: R -+ R is given 


by the formula 






fogQc) = f(g(x)) = 


\g(x)} 2 + 1 


x 6 

X 6 


-1 
+ r 



7.11 Inverse functions 

Suppose that A and B are sets and that / is a function from A to B. 
This means that each element a S A has a unique image 6 = /(a) S 5. 

We say that /"' is the inverse function to / if f" 1 is a function from B toA 
which has the property that x = f~\y) if and only if y = f(x). 

Not all functions have inverse functions. In fact, it is clear that a function 



Functions 



69 



f : A -*■ B has an inverse function /"' : B -> A if and only if each b £ B is the 
image of a unique a^A. (Otherwise /"' could not be a function). A function 
which has this property is said to be a 1 : 1 correspondence between A and B. 

In geometric terms, a function / : A ->B is a 1 : 1 correspondence between A 
and B (and hence has an inverse function f~ l : £ -* A) if and only if each vertical 
line through A meets the graph of / in one and only one point (which makes / a 
function) and each horizontal line through B meets the graph of / in one and 
only one point (which makes f' 1 a function). 




6= /(a) 
a =/■'») 



7. /2 Example Let /: (1 , ») -* (0, 1 ) be defined by 
x-l 



/(*) = 



x+ 1 



(* > 1). 



(0,1) 




(I.-) 



It seems clear from the diagram that / is a 1 : 1 correspondence between 
(1, ■») and (0, 1). Thus it has an inverse function. To prove this we must show 
that, given any y satisfying < y < 1 , there is a unique x > 1 which satisfies 



x-l 



y = 



(1) 



This is easily accomplished by solving (1) for*. We obtain 
y(x+ 1) = x-\ 



70 



Functions 



xiy-1) = -\-y 

1+2 

1-y- 



x = 



(2) 



Thus, if y satisfies <y < 1, there is a unique x > 1 which satisfies (1) — 
namely, that given by (2). We have established the existence of an inverse func- 
tion /"' : (0, 1) -* (1 , °°). It is given by the formula 



f\y) = 



i+y 

l-y 



Cve(o,i)). 



(!.«)< 




(O.n 



It is instructive to observe how the graph of x =f 1 (y) is related to that of 

y=m- 



7.13 Bounded functions 

Let / be defined on S. We say that / is bounded above on 5 by the 
upper bound H if and only if, for any x E S, 

m < h. 

This is the same as saying that the set 

f{S) = {f(x)-.xes} 

is bounded above by H. 

If / is bounded above on 5, then it follows from the continuum property 
that it has a smallest upper bound (or supremum) on S. Suppose that 

B = sup /(*) = sup f(S). 

xBS 



Functions 



71 




It may or may not be true that, for some % G S, /(£) = B. If such a value of % 
does exist, we say that B is the maximum of / on the set S and that this maxi- 
mum is attained at the point £. 




/attains a maximum of B at 
the point £ on the set S. 



Similar remarks apply to lower bounds and minima. If a function /is both 
bounded above and below on the set S, then we simply say that / is bounded on 
the set S. From proposition 2.3 it follows that a function / is bounded on a set 
5 if and only if, for some K, it is true that, for any x G S, 



l/(x)l < K. 



7.14 Example Let /: (0, °°) -* R be defined by 

m-\ (*>o). 

This function is unbounded above on (0,1]. It is, however, bounded below on 
(0, 1 ] and attains a minimum of 1 at the point x = 1 . 




ymfflk 







72 



Functions 



7.15 Example Let /: R ■* U be defined by f(x) = x. This function is 
bounded above on both the sets (0, 1) and fO, 1] and in both cases its supremum 
is 1 . But / has no maximum on the set (0. 1 ). There is no £ satisfying < | < 1 
for which /(£) = 1 . On the other hand, / attains a maximum of 1 at the point 
x= 1 on the set [0,1]. 




°^fy- 



'//7///////////A 



(0,1) 




7.16 Exercise 

(1) Draw a diagram illustrating the set of all (x, y) such that 
[5 if x > I 
2 if x < 1. 



y = 



Explain why this is a graph of a function from R to itself. What is the 
range of this function? What is the image of the set [1, 2] under this 
function? 

(2) Draw a diagram illustrating the equation 

\x\ + \y\ = 1. 

[Hint: consider each quadrant separately.] Explain why 
(i) the equation does not define a function from U to itself; 
(ii) the equation does not define a function from [— 1 , 1] to itself; 
(hi) the equation does define a function from [—1, 1] to [0, 1]. 

(3) Let /: [0, 1] -» [0, 1] be defined by 



/(*) 



1 -x 



(0<*<1) 



1 +x 

and let g: [0, 1] -*■ [0, 1] be defined by 

g(x) = 4x(l -x) (0<x<l). 

Find formulae for fa g and go f and hence show that these func- 
tions are not the same. 

Show that /"' exists but that g' 1 does not exist. Find a formula for 

r 1 . 

(4) The formula y = x 2 may be employed to define a function /: U •* R . 



Functions 



73 



Explain why /has no inverse function. If, instead, we use the formula 
y = x 2 to define a function g: [0, °°) -> [0, °°), show that g' 1 : [0, °°] -> 
[0, °°) exists and that 

g' , (y) = \ / y (y>o). 

(5) A function g: A -*■ B is a 1 : 1 correspondence between A and B. Prove 
that 

(0* _1 ° *(*) = * (xGA) 
(ii)gog- 1 (y)=y (y&B). 
To what do these formulae reduce when g is as in question 4? 

(6) Let / and g be bounded above on S. Let c be a constant. Prove that 
(i) sup {/(x) + c) = c + sup f(x). 

iGS xGS 

(ii) sup {/(x) + g(x)} < sup f(x) + sup g(x). 



xSS 



xes 



Give an example to show that equality need not hold in (ii). 



8 



1 



LIMITS OF FUNCTIONS 



Limits of functions 



75 



Given any e > 0, we can find a 5 > such that 

|/(;t)-/|<e 

provided that a < x < a + 5 . 



8.1 Limits from the left 

Suppose that /is defined on an interval (a, b). We say that/(x) tends 
(or converges) to a limit / as * tends to b from the left and write 

f(x) -*■ I as x •* b — 
or, alternatively, 

lim f(x) =1 

x -* b- 

if the following criterion is satisfied. 




Given 


any e > 0, 


we can 


find i 


1 6 > such that 


\m 


-/|<e 








provided that b - 


-8<x 


<b. 





The number \f(x) — 1\ is the distance between /(x) and /. We can think of it 
as the error in approximating to / by f(x). The definition of the statement 
fix) -* / as x -* b — then amounts to the assertion that we can make the error in 
approximating to / by f(x) as small as we like by taking x sufficiently close to b 
on the left. 



8.2 Limits from the right 

Suppose again that /is defined on an interval (a, b). We say that/(X) 
tends (or converges) to a limit / as x tends to a from the right and write 

f(x) ■* / as x -> a +, 
or, alternatively, 

lim f{x) = I 

x -* a+ 

if the following criterion is satisfied. 
74 




8.3 f{x) •* I as *->•§ 

Suppose that /is defined on an interval (a, b) except possibly for some 
point % G (a, b). We say that fix) tends (or converges) to a limit / as x tends to % 
and write 



f(pc) -*lasx-+% 
or, alternatively, 

if the following criterion is satisfied. 




Given any e > 0, we can find a 


5 > such that 


\fix)-l\'<e 






provided that < \x - 


-IKS. 





If a and are real numbers, it is often useful to note that the inequality 
\a\ < (3 is equivalent to — < a < (3, or, what is the same thing, - (3 < — a < 
(see exercise 1.20(1)). 

Thus, in the definitions above, the condition |/(x) — /| < e can be replaced 
throughout by — e < fix) — / < e or, alternatively, by — e < / —fix) < e. 

Similarly, the condition < \x — %\ < 5 in the last definition is equivalent to 
the assertion -S<j:-?<6andx^|. Thus to say that < \x - 1| < 5 is to 
say that x satisfies one of the two inequalities £— 8<x<%or%<x<% + 8. 

With the help of the last remark it is easy to prove the following result. 

8.4 Proposition Let /be defined on an interval (a, b) except possibly at a 
point % e (a, b). Then fix) -> / as x -> % if and only if fix) -* I as x -*■ % — and 
/(x)->/ as *-*£+. 



76 Limits of functions 

8.5 Example Let /be the function from R to itself defined by 



m = 



1 -x (x<\) 



,2x 



(x>\). 




Then 



(i) lim f(x) = (ii) lim f(x) = 2. 



Note that it follows from proposition 8.4 that 



ton f(x) 

does not exist even though /(l) is perfectly well defined (/(l) = 0). 

Proof (i) f(x) •* as x -*■ 1 — . Given any e > we must show how to 
find a 5 > such that 

|/(x)-0|<e 

provided that 1 — 5 <x < 1 . Since we are only concerned with values of x 
satisfying x < 1 , we can replace f(x) by 1 -*. The condition \f(x) - 0|< e 
then becomes 1 1 — x \ < e which is equivalent to — e <x — 1 < e. Adding 1 
throughout, we see that | f{x) - |< e is the same as 1 - e < x < 1 + e. 

The problem is now reduced to the following. Given any e >. 0, find a S > 
such that 

1 -e<x<\ +e 

provided that 1 — 6 < x < 1 . 

Obviously, the choice 6 = e suffices to make this true and this completes the 
proof that f(x) -*■ as x -> 1 — . 

(ii) f(x) -*■ 2 as x •* 1 +. Given any e > 0, we must show how to find a 5 > 
such that 

l/W-2|<e 

provided that 1 < x < 1 + 6. Since we are only concerned with values of x 
satisfying x > 1 , we can replace f(x) by 2x. The condition \f(x) - 2|< e then 
becomes \2x - 2| < e which is equivalent to — e < 2(x — 1) < e. Thus 
l/(x) - 2|< e is the same as 1 - ^e < x < 1 + |e. 

The problem is now reduced to the following. Given any e > 0, find a 5 > 
such that 

1 -|e<*<l +\e 

provided that 1< x < 1 + 6. 



Limits of functions 



77 



Obviously, the choice 5 = ^e suffices to make this true and this completes the 
proof that f(x) -* 2 as x ■* 1 +. 



8.6 Continuity at a point 

In the definition of/(x) -> / as x ■* % given in §8.3, consideration of 
what happens when x actually equals % was carefully excluded. When taking 
limits we are only interested in the behaviour of/(x) asx tends to \ not in what 
happens when x equals %. 

Thus it is quite possible for/(x) to tend to a limit / as* tends to % even 
though f(x) is not defined at the point % (see exercise 8.15(3) below). And, even 
if f(X) ' s defined, it is not necessarily true that /=/(?) (see exercise 8.15(2) 
below). 

Having said this, we can now turn to the definition of continuity at a point. 

Suppose that /is defined on an interval (a, b) and % e (a, b). Then we say that 
/is continuous at the point % if and only if 

/(*)-/(?) as *->£. 

Roughly speaking, to say that /is continuous at the point % means that the 
graph of/ does not have a 'break' at the point ff. 

If /is defined on an interval (a, b] and f(x) •* f(b) »x-*b— »we say that /is 
continuous on the left at the point b. If /is defined on an interval [a, b) and 
f(x) -*f(a) as x -> a +, then we say that /is continuous on the right at the point 
a. 




f continuous at j-. 



f(b)---/-\-z- 




/ continuous on 
the left at b. 



/(a)-- 




/continuouson 
the right atfl. 



8. 7 Example For any real numbers a and 0, the function /: defined by 

/(*) = ax + 

is continuous at every real number |, 

Proof Assume a # 0. (If a = 0, the proof is even easier.) Let e > be 
given. Choose 5 = e/|a|. Then, provided that \x — %\ < 5 , 

I/(X)-/(!)I = N*-f)l = |a|.|x-$|<|a|.8 = e. 



78 Limits of Junctions 

Given any e > 0, we have found a 5 > such that |/(x) — /(£)| < e provided 
that \x — 1| < 5 . Hence f(x) -> /(£) asx -»g and so /is continuous at the point g. 

Note that we showed that | /(.x) — /(£) I < e 'provided that | x — g i < 5' 
instead of 'provided that 0<|.x — g|<5as appears in the definition of f(x) -* / 
as x -»• g- This is because, when x = g, it is automatically true that | /(jc) — /(g) I < e 
because then | f(x) — /(g) | = 0. It is only when we are considering a limit / =£/(g) 
that it is important to exclude the possibility that x = g. 



8.8 Connexion with convergent sequences 

8.9 Theorem Let /be defined on (a, b) except possibly for g € (a, b). Then 
f(x) •* I as x •* g if and only if, for each sequence Cx n > of points of (a, b) such 
that x n =£ | (n = 1 , 2, . . .) and *„ -> g as n •* °°, it is true that f(x n ) -*■ I as n -*■ °°. 

Proof (i) Suppose first that f(x) ■* I as x •* g. Let e > be given. Then 
we can find a 5 > such that \f(x) — l{<e provided that < |x — g| < 6 . Now 
suppose that <x n > is a sequence of points of (a, b) such that x n ¥= g (n = 1, 2, . . .) 
and x n ■+ 1 as « -> °°. Since 5 > we can find an iV such that for any n>N, 
\x n - %\< 5. But x n ¥= g (n = 1 , 2, . . .) and so < \x n - g|<5. But this implies 
that 

l/W-'Ke. 

Given any e > 0, we have found an N such that, for any n>N, 
|/(x„) - /|< e. Thus f(x„) -* / as « -* °°. 

(ii) Now suppose that, for each sequence (x n ) of points of (a, b) such that 
x n & % (» = 1 > 2, . . .) and jc„ -*■ g as « -* oo > it is true that f(x n ) •* I as n -* °°. 
We now suppose that it is not true that/(x) •* / as x -» g and seek a contra- 
diction. 

If it is not true that /(at) ■+ / as x ■* g then, for some e > 0, it must be true 
that for each value of 5 > we can find an x satisfying < |.x — g| < 6 such that 

\f(x)-l\>e. 

In particular, if S = 1/ra, then we can find an x n satisfying < \x n — g| < 1/n 
such that 

\f(x n )-H>e. 

But then U„) is a sequence of points of (a, b) such that x n ^ | (« = 1, 2, . . .) 

and x n -> I as n -*■ °° but for which it is not true that f(x„) -*■ I as n -* °°. This 
contradicts our assumption above. 

Similar results to theorem 8.9 hold for convergence from the left or right. For 
example, fix) -> / as x -» Z> — if and only if, for each sequence Cx n > of points of 
(a, 6) such that x n ■+ 6 as « -» °°, it is true that /(x n ) ->■ / as n -*■ °°. 






Limits of functions 



79 



8.10 Example We show how theorem 8.9 can be used to prove that the 
function /: R -* R defined by f(x) = ax + (3 is continuous at every point g (see 
example 8.7). 

Let (jc n ) be any sequence of real numbers such that x n ■* g as n ■* <*>. By 
proposition 4.8, ax n + ■* «| + P as n -» °°. From theorem 8.9 it follows that 
ax + ■* ag + p as x -* g and this concludes the proof. 



y=/te>=«X+/ 




8.11 Properties of limits 

The following propositions are analogues of the combination theorem 
and the sandwich theorem of chapter 4. They are easily deduced from these 
results with the help of theorem 8.9. 

8.12 Proposition Let /and g be defined on an interval (a, b) except possibly 
at g G (a, b). Suppose that f(x) ■* I as x ■* g and g(x) ■* m as x -* g and suppose 
that X and p are any real numbers. Then 

(i) \f(x) + pg(x) ^\l + pmasx^H 

(iii) f(x)lg(x) -*■ l/m as x -> g (provided m =£ 0). 

An important consequence of proposition 8.12 is the following result which 
we quote as a theorem. 

8.13 Theorem A polynomial is continuous at every point. A rational func- 
tion is continuous at every point at which it is defined. 

Proof Example 8.7 shows that x -» g as x •*■ g. Repeated application of 
proposition 8.12(ii) then shows that x h -> g* as x •* g for each keN. 
Hence, if 

P(x) = a n x n + a n - 1 x"- 1 + . .. + a,x + a , 

then a repeated application of proposition 8.1 2(i) yields P(x)^P(& asx -> g, 
i.e. Pis continuous at g. 

To show that a rational function is continuous wherever it is defined we 
appeal to proposition 8.12(iii). 



80 



Limits of functions 



8.14 Proposition (sandwich theorem) Let f,g and /; be defined on (a, b) 
except possibly at | e (a, b). Suppose that g(x) -*• / as x •*■ f , h(x) •* I as x -* % 
and that 

g(x)<f(x)<h(x) 

except possibly when x = %. Then/(x) -*■ I as x -*■ £. 

Note The analogues of propositions 8.12 and 8.14 for convergence 
from the left or right are both true. 



15 Exercise 


i 


(1 ) Calculate the following limits 


ix 2 + 4) 
(i) lim (-= — - (ii) lim 
x-*i [x 2 — 4 J '*-*o 


(2) Let/: R -► R be defined by 




3-x (*>1) 


m = 


1 (x = 1) 




2x (x<l). 



,73 



+ 5x* 2 + 9 



3x 2 



Show that /(at) -*■ 2 as x -> 1 - and f(x) -* 2 as x -> 1 + using only the 
definitions of §8.1 and §8.2. Deduce that the limit 

lim f{x) 

X -* 1 

exists but is not equal to/(l). Draw a graph. 

(3) Let /be defined for all x except x = by 

».fl±£=I <^0). 

Prove that f(x) -» 2 asx -» even though /(0) is noj defined. 

(4) Use the definition to show that |x — £| -* as x -> £. A function /is 
defined on an open interval which contains % and, for each x in this 
interval, /(x) lies between £ and x. Prove that/(x) -> £ as x -*■ %. 

(5) Let n £ N . Prove that the function / : [0, °°) -» R defined by 

f(x) = x Vn (x>0) 

is continuous at each £ > and continuous on the right at 0. [Hint: 
Use the sandwich theorem and exercise 3.1 1(2) to prove that/is con- 
tinuous at £ > 0. For continuity on the right at 0, appeal to the defi- 
nition of f(x) -*■ as x -* +.] 

(6) Suppose that f(x) ■* I as x -> £. If / > 0, show that, for some h > 0, 



Limits of functions 



81 



f(x)>0 

provided that £ - /; < x < £ + /i and x # £. What happens if (i) / < 0, 

(ii) I - 0? 



8.16 Limits of composite functions 

Suppose that f(y) -* / asy -* 7? and that g{x) ■* 77 as x -» £. It is tempting 
to write y = g(x) and conclude that 

But some care is necessary. Consider, for example, the functions /: R -> R 
and g: R -*■ R defined by 



m = 



3 = 1) 
2 0*1)' 



g(x) = 1. 



Then /(j>)-»- 2 as y -* 1 andgXx)-* 1 as x -+0. But it is not true that f(g(x))->- 2 
as x -* because, for all values of x,f(g(x)) =/(l) = 3. 

& 7 7 Theorem Suppose that f(y) -* I as y -» tj and that £(x) ->■ 77 as x ■* |. 
Then either of the two conditions below is sufficient to ensure that 

/&(*))-»/ as *-»•$. 

(i)/is continuous at 77 (i.e. / = /(t?)). 

(ii) For some open interval / containing £, it is true that g(x) # 77 for any 
x S 7, except possibly x = £. 

/Voo/ Let e > be given. Since f(y) -> / as y •*■ tj, we can find a A > 
such that I/O) - /|< e provided that < \y - 77I < A. 

Writing v =g(x), we obtain 

|/fe(x))-/Ke (0 

provided that 

< \g(x) - 7?|< A (2) 

But £(x) -* 7? as x -* £ and A > 0. Hence we can find a 6 > such that 

\g(x)-v\<A ( 3 ) 

provided that 

0<|.x-£|<5. ( 4 ) 

We would now like to say that (4) implies (3), which implies (2), which in 
turn, implies (1). This would complete the proof. But notice that (3) does not 



82 



Limits of functions 



imply (2) in general. We therefore need to use one of the hypotheses (i) or (ii) of 
the theorem. 

If we assume that/is continuous at 77, then / = /(rj) and so \f(y) —l\<e even 
when y = r\ (see the remarks at the end of example 8.7). Thus, in this case, we 
may replace condition (2) by condition (3) and the argument described above 
goes through. 

If instead we assume hypothesis (ii), then we can be sure that£(je) ^17 pro- 
vided that x satisfies (4) for a sufficiently small value of 5 >0. But then con- 
dition (3) can be replaced by condition (2) and the argument goes through again. 

8.18 Divergence 

We write down two sample definitions. The reader will have little diffi- 
culty in supplying the definitions in other cases. 

We say that/(x)->--l- °°as*-> £+ if, given any /f > 0, we can find a 5 >0 
such that 

provided that | <x < g + 6. 

We say that f(x) ■* / as x -*■ — °° if, given any e > 0, we can find an X such 
that 

|/(x)-/|<e 

provided that x < X. 




f{x)—+->*sx^'i + 




f(x)-'l as x-*- 



8.19 Example Prove that x l -* + °°asx->0+. 

Proof Let H > be given. We have to find a 5 > such that 

1 

~>H 

x 

provided that <x < 8 . Obviously the choice 8 = H' 1 satisfies the requirement. 



Limits of functions 83 

120 Exercise 

(1) Prove the following 

(i) x' 1 -*■ — °°asx->0 — 
(ii) x 7, -* + °° as x -* + °°. 

(2) Suppose that f(y) ■* I as y ■* + « a nd g(x) ■* + °° as x ■* + «. Prove 
that/(g(jc)) -»■ / as x ■* + °°. Explain why the difficulties encountered in 
theorem 8.17 do not occur in this case. 

(3) Suppose that f(y) -+ 1 as y ■* +. Prove that fix' 1 ) -+l asx-> + °°. 

(4) Suppose that /is defined on (a, b) except possibly for | G (a, b). Let 
(x n ) be a sequence of points of (a, b) such that x n i= % (n = 1 , 2, . . .) 
and x n -*■ % as n -* °°. Let (y„) be a sequence with the same properties. If 
/ ¥= m and /(*„) -*■ / as n ■* °° but /(y n ) -*■ m as M -* °°, show that 

does not exist. [Hint: theorem 8.9] 

(5) Let/: U^U be defined by 

(1 (x rational) 
(x irrational). 
Show that 
lim f(x) 

x -* 

does not exist. 

(6) If /is defined as in the previous question, prove that 

iim {*/(*)} = 0. 

I-»0 



CONTINUITY 



9.1 Continuity on an interval 

A function / is continuous at a point % if and only if f(x) -*/(£) as 
x •* | (see §8.6). Geometrically, this means that the graph of / does not have a 
'break' or 'jump' above the point £. 

To say that / is continuous on an interval / should mean that its graph is 
unbroken above the interval /, i.e. that we can draw the part of the graph which 
lies above /without lifting our pencil from the paper. 

A precise mathematical definition is required. This depends on the type of 
interval in question. 



A function / is continuous on an open interval / if and only if it 
is continuous at each point of/. 



Where an endpoint of an interval belongs to the interval, a slightly more com- 
plicated definition is required. The most important case is that of a compact 
interval [a,b]. 



A function / is continuous on a compact interval [a, b] if and 
only if it is continuous at each point of (a, b) and continuous on the 
right at a and on the left at b. 




/continuous on ia.b) 




-^ mmt&i 



/continuous on \a,b) 



84 



Continuity 



85 



Note that, in both diagrams, the graph of /is unbroken above the interval in 
question. 

In general, a function / is continuous on an interval / provided that: 

(i) it is continuous at each point of/ which is not an endpoint; 

(ii) it is continuous on the right at the left hand endpoint of/, // this exists 
and belongs to /; 

(iii) it is continuous on the left at the right hand endpoint of/, //"this exists 
and belongs to /. 



9.2 Examples 

(i) All polynomials are continuous on every interval. All rational func- 
tions are continuous on any interval not containing a zero of the denominator 
(theorem 8.13). 

(ii) Let /: (0, ■»)-»• R be defined by f(x) = 1/jc. Then / is continuous on the 
interval (0, °°) because it is continuous at each point % satisfying £ > 0. 

(iii) Let /: U -+R be defined by 

(x 2 (jc<1) 

\ 3/2 (*>1). 

Then / is continuous at every point except x = 1 . But / is continuous on the 
left at x = 1 and thus /is continuous on the interval [0,1]. It is not continuous 
on [1,2]. 




9.3 Proposition Let / be defined on an interval /. Then / is continuous on 
/ if and only if, given any x G / and any e > 0, we can find a 5 > such that 



\f(x)-f(y)\<e 
provided that y € /and satisfies \x —y\ <8. 



86 



Continuity 



9.4 Proposition Let X and n be real numbers and suppose that the func- 
tions /and g are continuous on an interval /. Then so are the functions 

(i)X/+W? 

PDA 

(iii) fig (provided g(x) ¥• for any x e /). 

9.5 Proposition Let g : / -*■ J be continuous on the interval / and let 
/: J-*R be continuous on the interval /. Then / o g is continuous on /. 

9.6 Proposition Let /be continuous on the interval /. If £6/ and (x n ) is a 
sequence of points of / such that x n -* if as n -*■ °°, then 

These four propositions are easily proved. The first is proved by an appeal to 
the limit definitions of § §8.1, 8.2 and 8.3. The second follows from proposition 
8.2, the third from theorem 8.17 and the fourth from theorem 8.9. The last of 
these results can be remembered in the form 

lim /(*„) = /(lim x n ) 

i.e. for a continuous function the limit and function symbols can be validly 
exchanged. (We say that the symbols 'commute'.) 



9.7 Continuity property 

We now come to a theorem which may seem so obvious as to be hardly 
worth mentioning. It is, however, of a fundamental importance comparable to 
that of the Continuum Property of §2.2 and will serve as a foundation stone in 
our development of the calculus in the next few chapters. 

9.8 Theorem (continuity property) 

Let /be continuous on a compact interval [a, b] . Then the image of 
[a, b] under /is also a compact interval. 



\c.d 




\a.b\ 



Continuity 



87 



We split the content of the continuity property into three lesser theorems and 
prove these separately. Given that /is continuous on 1 = [a, b] , it follows from 
theorem 9.9 that /=/(/) is an interval. From theorem 9.1 1 it follows that the 
interval /is bounded and from theorem 9.12 that J includes its endpoints. 

9.9 Theorem Let /be continuous on an interval /. Then the image of / 
under/is also an interval. 

Proof To show that J =f(I)={ f(x) : x € / } is an interval , we need to 
show that, if y x andy 2 are elements of J and j>, < X <y z , then X EJ (see §2.9). 
Since y i £ J and y 2 e J, the subsets of / defined by 

S = {x:f(x)<\}; T = {x:f(x)>\} 

are non-empty. Also, every point of the interval /belongs to one or other of 
these sets. It follows that a point of one of the sets is at zero distance from the 
other (exercise 2.13 (6)). Suppose that s E 5 is at zero distance from T. Then a 
sequence < t n ) of points of T can be found such that t n -*■ s as n -* °° (exercise 
4.29(6)). Since/is continuous on / it follows fhat/(r„) -»/(s) as n -* °° (pro- 
position 9.6). But 

f(t n )>\ (n = 1,2,...) 

and therefore /(s) > X (theorem 4.23). We already know that f(s) < X and so it 
follows that /(s) = X. Hence X EJ. 

A similar argument applies if a point of T is at zero distance from S. 

9.10 Corollary (intermediate value theorem) Let /be continuous on an 
interval / containing a and b . If X lies between f(a) and f(b), then we can find a 
% between a and b such that X = /(£). 



fib) 




Proof The corollary is simply a somewhat clumsy restatement of 
theorem 9.9. 

9.11 Theorem Let /be continuous on the compact interval [a, b]. Then /is 
bounded on [a, 6]. 

Proof Suppose that /is unbounded on [a, b]. Then we can find a 



88 



Continuity 



Continuity 



89 



sequence (x n ) of points of [a,b] such that 

|/(x B )I-*- + 0O as»-+« (1) 

(see exercise 4.29(6)). Since [a, b] is compact, there exists a subsequence <x„ r > 
.which converges to a point £ £ [a, b] by the Bolzano-Weierstrass theorem (see 
exercise 5.21(4)). Because /is continuous on [a, b\ ,/(x„ r ) -*/(£) as r ->■ °° (pro- 
position 9.6). But this contradicts (1). 

9.72 Theorem Let /be continuous on the compact interval [a, b]. Then/ 
achieves a maximum value d and a minimum value c on [a, b] . 

Proof We know from the previous theorem that /is bounded on [a, b]. 
Let <2 be the supremum of /on [a, b] and consider a sequence < X n ) of points of 
[a, b] such that f(x n ) -*■ d as « -* °° (see exercise 4.29(6)). Since [a, b] is com- 
pact, (x n ) contains a subsequence <x„ r > which converges to a point * S [a, &], 
Because /is continuous on [a, b], it follows that /(x„ r ) -*/(£) as/--* <*>. Hence 
^•(H) = ^ and thus the supremum d is actually a maximum. 
A similar argument shows the infimum to be a minimum. 



9.13 



Example The function/: R -* R defined by 
5 (x>l) 



/(*) = 



2 (*<1) 



is not continuous on [0,2]. Observe that 2 = /(0) < 4 </(2) = 5 but there is 
no value of £ between and 2 such that /(£) = 4. Note that, for X satisfying 
2<X<5, 

{x:f(x)<\} = {2}; {x:f{x)>\} = {5}. 

9.14 Example The function/: (0, °°) ■* R defined by /(x) = 1/x is con- 
tinuous on the open interval (0, 1) but is not bounded on (0, 1). The function 
g : R ■* R defined by g(x) = x is continuous everywhere and happens to be 
bounded on the open interval (0, 1). But g does not attain a maximum value or 
a minimum value on (0, 1). 

9.15 Example Show that the equation 

17x 7 -19a- 5 -1 = 

has a solution % which satisfies — 1 < | < 0. 

Proof The function/: R-*R defined by /(x) = 17x 7 - 19.v 5 - 1 is a 
polynomial and hence is continuous everywhere. In particular it is continuous on 
[-1,0]. Now/(- 1) = 1 and /(0) = - 1 . Since - 1< < 1 , it follows from 



corollary 9.10, that/(£) = for some 
| between — 1 and 0. 







9.16 Example Let/: [a, b] 
some £ G [a, ft], 

m) = t 



[a, b\ be continuous on [a, b] . Then, for 



Thus a continuous function from a compact interval to itself 'fixes' some 
point of the interval. (This is the one-dimensional version of Brouwer's fixed 
point theorem.) 

Proof The image of [a, b] is a subset of [a,b]. Thus f(a) > a and 
fib) < b. The function g : [a, b] •* U defined by g(x) = f(x) — x is continuous 
on [a, b] by proposition 9.4. But g(a)>0 and g(b) <0. By corollary 9.10, for 
some £ e [a, b] it is true that£(£) = 0. Thus/(£) = %. 




9.17 Exercise 

(1) Each of the following expressions defines a function on (— 2, 2). 
Decide in each case whether or not the function is continuous on (a) 
(— 2, 2) and (b) [0, 1 ] . In each case draw a graph. 



(i) /(*) = 



2x4-3 
2x-5 



(ii) fix) = \x-l\ 



90 



Continuity 

(1 (*<1) (1 (0<x<l) 

(iii) f(x) = { (iv) fix) - j 

\2 (x>l) iO (otherwise) 



(v) /(*) = 



* 2 + 4 

x 2 -4 



(vi) /(*) = 



x(x-\) (0<*<1) 
(otherwise) 



Which of these functions are bounded on (- 2, 2)1 Which attain a 
maximum on (— 2, 2)? Which attain a minimum on (— 2, 2)? 

(2) A function /is continuous on an interval / and for each rational number 
r £ / it is true that 

f(r) = r\ 

Prove that/(x) = x 1 for any xE.1. [Hint: proposition 9.6.] 

(3) Show that all polynomials of odd degree have at least one (real) root. 

(4) Suppose that / is continuous at every point and that f(x) -> as 

x -» + oo and/(jc) -*■ as x -» — °°. Prove that /attains a maximum 
value or a minimum value on the set U. 

(5) Let / be continuous on the compact interval /. Suppose that, for each 
x £ /, there exists a y £ / such that 

l/OOKil/001- 

Prove the existence of a % £ / for which /(£) = 0. [Hint: Bolzano- 
Weierstrass theorem.] 

(6) Let / be a closed interval and let < a < 1 . Let /: / -*■ / satisfy the 
inequality 

\ax)-f(y)\<a\x-y\ 

for each x £ / and y £ /. (Such a function is called a contraction map- 
ping). Prove that /is continuous on /. 

Let x, £ / and define x n+l - /(*„) (n = 1 , 2, . . .). Prove that the 
sequence <x n > converges and that its limit / satisfies /=/(/). [Hint: 
exercise 5.21(1).] 



10 D 



IFFERENTIATION 



10.1 Derivatives 

Suppose that /is defined on an open interval /containing the point £. 
Then /is said to be differentiable at the point £ if and only if the limit 

,. /(*)-/(£) 

lim z — 

*-** x-% 

exists. If the limit exists, it is called the derivative of /at the point % and denoted 
by/'(£) or £>/«). 

For a function /which is differentiable at % we therefore have 



/'(I) = Hm, 



m-rm 



■ i x — | 

Equivalently, we may write 

ffi + h)-f® 



fm - Hm 



h 



(1) 



From an intuitive point of view, to say that a function /is differentiable at % 
means that a tangent can be drawn to the curve y = fix) where x = £. The slope 
or gradient of this tangent is then equal to /'(£)• This geometric interpretation is 
based on the diagram below. 



/(A) 



r-/(£> + /(.v {) 




91 



92 



Differen tiatiun 



In order that the line y = /(?) + l(x — |) be tangent to y = fix) where x — % t 
we require that the slope of the chord PQ approach the slope / of this line as 
x -*■ £. The slope of the chord PQ is 

qr _ m-m 

PR x-% 

and hence our requirement reduces to 

lim f^lM . , 
*•** X— | 

as in the definition of differentiability. 

(Note: It follows immediately from our definition that a function/is differen- 

tiable at ij if and only if there exists an / such that 

m + 1') -f® - if' = o(h) (h -* o). (2) 

Here o(h) denotes a quantity which tends to zero when divided by h. This form 
of the definition lends itself more readily to generalisation.) 



10.2 Higher derivatives 

If /is differentiable at each point of an open interval /, we say that/is 
differentiable on /. In this case it is natural to define/' or Df to be the function 
from / to R whose value at each point x £ / is f'(x). We can then define the 
second derivative /"(£) or D 2 f(g) at the point % by 



f m = I™ 



/'(*)-/'(*) 



1 x-% 

provided that the limit exists. Similarly for third derivatives and so on. 

10.3 Examples 

(i) Let/: R->R be defined by f(x) = x 2 . Then 

/'(*) = Dx 2 = 2x 
for all values of x. 

Proof We have 

f(x + h) -f(x) (x + hf -x 2 2xh+ h 2 



h 



h 



h 



= 2x + h -*■ 2x as h -*■ 0. 



(Note that we studiously ignore what happens when h = - see §8.4. It is as well 
that this is permissible, for the expression 

f(x + h)-f(x) 



Differentiation 



93 



is quite meaningless when h = 0.) 

(ii) Let/: (0, <*>) -* R be defined by f(x) = 1/x (x > 0). Then 

fXx)=D l -=-± 

for all values of x > 0. 
Proof We have 

f(x + h)-f(x) \_ 
h h 



1 



1 



-h 



h \x(x + ft) 



I 

x + h x 

- — * -a as h -*■ 0. 

x(x + h) x 2 



10.4 More notation 

We shall avoid the notation introduced in this section because, when 
using this notation, it is difficult to make statements with a precision adequate 
for our purposes. In many applications, however, it is unnecessary to maintain 
this level of precision. In such cases, the notation described below can be a very 
useful and powerful tool and it would be pedantic to deny oneself the use of it. 
Replace £ by x and h by 5x in formula (1) of § 1 0.1 . Then 



/'(*) = 



lim 
&x -• o 



f(x + 8x)-f(x) 
8x 



= lim — 
Sx - o S* 



(3) 



If>> =f(x), the quantity 5^ =f(x + 8x) —f(x) is usually said to be the 'small 
change in y consequent on the small change 8x in*'. 

From formula (3) it is only a short step to the much used (and commonly 
abused) notation 

This notation has the drawback that it uses the symbol x ambiguously. It is used 
simultaneously as the point at which the derivative is to be evaluated and the 
variable with respect to which one is differentiating. In applications where this 
distinction is unimportant, the ambiguity creates no problems. But, for the 
material of this book, the distinction is important. 

Further confusion is possible when 'differentials' are introduced. The differ- 
ential df of a function /may be regarded as the function of two variables given 
by 

df(x;h) = /'(*)//. (4) 

Formula (2) can then be rewritten as 



94 Differentiation 

f{x +h)- f{x) -dfix-h) = o in) ih ■+ 0) (5) 

provided that /is differentiable at*. 
For a fixed value of*, the equation 

k = dfix-h) = f\x)h (6) 

is the equation of a straight line. If the h and k axes are drawn as below, (6) is 
the equation of the tangent to the graph of the function /at the point*. 




A" =/'(.v)/i 



The reader may justifiably feel that little or nothing has been added to our 
understanding by the introduction of the idea of a differential, the idea being 
more useful when functions of several variables are being considered. Our only 
reason for discussing differentials is to explain the meaning of the equation 

dy = f'ix)dx. 

This is simply equation (6) with the variable h replaced by dx.and the variable 
k replaced by dy. Thus dx and dy should be thought of as variables. The reason 
one chooses this notation is to make the formula 



dy = -fdx 
dx 



(7) 



a valid one. 

It is false that dy denotes 'the small change in y consequent on a small change 
dx in *'. Formula (5) tells us that this is nearly true for small values of dx but it 
will only be exactly true for functions /with a straight line graph. Should it be 
necessary to discuss 'small changes in * and >>', the notation 5* and 8y is 
available and one can say that 8y is approximately equal to f'ix)bx. 

Even more false (if that is possible) is the idea that dy and dx are somehow 
'infinitesimal quantities' obtained by allowing 5* and 8y to tend to zero. One 
would then regard 

dy 

dx 



Differen tia tion 



95 



as the quotient of these quantities and abandon all the apparatus of limits. This 
metaphysical notion is admittedly an attractive one and it is on this idea that 
Newton based his 'Theory of Fluxions'. Unfortunately it cannot stand up to 

close logical scrutiny. 

(Note: A common error is to confuse the terms derivative and differential. A 
derivative is sometimes referred to as a 'differential coefficient' (because of (7) 
above) but should not be called a differential.) 

10.5 Properties of differentiable functions 

10.6 Theorem Let /be defined on an open interval / which contains the 
point £. If /is differentiable at £, then /is continuous at %. 

Proof We have 

/(*)-/(£) 



fix) -m) = 



x-% 



(*-£W(S)-0as*^. 



Hence /(*) -*/(£) as x -*■ | and the proof is complete. 



10. 7 Example Let /: U •* U be defined by 
[1 (x > 1) 
-1 (*<1). 



m = 



We can see immediately that /is not differentiable at the point 1 because it is 
not continuous there. 



10.8 Example Let/: R -> R be defined by 



/(*) = 



* ix>\) 
x 2 ix<\). 



Then /is continuous at 1 but not 
differentiable (The graph has a 'corner'). 
We have 




Iim ; — lim ; — 

h->0+ h h--0+ h 



= 1 



r /0+^)-/(l) ,. (1+/Q 2 -1 . 

lim ; = lim ; = 2. 

h-»o- h h->o- h 



Thus the converse of theorem 10.6 is false. 



96 



Differentiation 



10.9 Theorem Suppose that /and g are defined on an open interval /con- 
taining the point %. Let X and \x be any real numbers. Then, if /and g are 
differentiable at £, 

(i) £>{X/+WT} = W + liDg 
(ii) D{fg) = fDg + gDf 

(iii) D j4= gJ^Zgte (provided *(g) * ) 

at the point f. 
/Voo/ 
1 



0) J (X/(? + A) + «?(! + h) - \f® - WG)} 



= X 



m+h)-m 



h 



+ H 



g(M+m-g® 



(«) 






A 



1 



= jf {/(i + am* + a) -/ct)r(i + h) +m)ga + h) -/«)*«)} 



= *ci + /o 



m+h)-m) 



h 



+/«) 



g (S+*)-g( l) 

h 



(Note that£(i; + A) ->-£•(£) as A -* because, by theorem 10.6, g is continuous 
at?.) 

m 1 ( flt+*) ftm _ /g+ft)gg)-g(s+ft)/(S) 

= » L m m+h)-m _ m g(.i+h)-m 



i 



fed)} 5 



{*«)/'(*) -/«)*'(S)} as A -0. 



70.70 Example If « is a natural number, then, for any x, 
Dx" = nx"-\ 

Proof This is obvious when n = 1 since 
(x + h)-x 



= 1 -*■ 1 as A •* 0. 



Differentiation 97 

Now assume its truth when « = k. Thus Z>.x fe "fee*" 1 . Using theorem 1 0.9 (ii), 

D(x h + l ) = D(x h .x) = x h .l +x.kx k - 1 = (A+l)x fc 
and the result follows by induction. 



10.11 Exercise 

(1) Prove that, for any x. 



D 



1 



1 +x' 



-2x 
(1 +xy 



(2) If n is a negative integer and .v 9* 0, prove that jDx" = nx n ~ 1 . 

(3) Let/: RHR be defined by 



/(*) = 



2x (x>l) 
x 2 +1 (x< 1). 



Prove that /is differentiable at the point 1 and has derivative 2. 
Let g: R ■* R be defined by g(x) =\x\. Prove that g is not differen- 
tiable at the point 0. [Hint: Consider the right and left hand limits 
separately.] 

(4) A polynomial P of degree n has the property that />(£) = and 
/>'(g) = 0. Prove that 

P{x) - (pc-tfQ(x) 

for all x, where Q(x) is a polynomial of degree n — 2. (See exercise 
3.11(3).) 

(5) Let/: R-*R be n times differentiable at the point g. Show that the 
'Taylor polynomial' P defined by 

pw = f® + ^rfm + ^=rVa) + • ■ ■ 

+ (•*-£)" fCn-iy 



(« - 1)! 



-/»-»(|) 



has the property P*\|) = / w (g) (* = 0, 1 , 2, . . . , n - 1 ). (Note that 
/ <fc) (l) denotes the derivative of order k at the point jf.) 
^(6) Let/and,? be n times differentiable at the point %. Prove 'Leibniz's 
rule', 

i.e. D n fg = t ("W/EP-'f 



;=o \/ 



at the point g. [Hint: Recall the proof of the binomial theorem.] 



98 Differentiation 

10.12 Composite functions 

The next theorem is usually remembered in the form 

dz dz dy 
dx dy'dx 

10.13 Theorem Suppose that g is differentiable at the point x and that/is 
differcntiable at the point y = g(x). Then 

(f°g)'(x) = f'(y)g'(x). 

Proof Recall thatfog(x) =f(g(x)). We therefore have to consider 

f(g(x+h))-f(g(x)) 
h 

Write k = g(x +h) —g(x). Since g is differentiable atx, it is continuous there 
and so k -*■ as h -*■ 0. Suppose that k=£0. Then 

f(g(x + h))-f(g(x)) _ f(gix+h))-f(g(x)) g(x+h)-g(x) 
h g(x + h)-g(x) ' h 

__ f(y + k)-f(y) g(x+h)-g(x) 
k h 

It is tempting to deduce the conclusion of the theorem immediately. But care 
is necessary. What happens if, for some values of h, k = g(x +h)—g(x) = 0? To 
get round this difficulty we introduce the function 

(fjy + k)~f(y) 

F(k) - ( 

(* = 0). 



k 

f'(y) 



(**0) 



Since /is differentiable sty, F(k) -+f'(y) as fc ->- 0. Because F(0) =f'(y), it 
follows that F is continuous at the point 0. From theorem 8.1 7(i) we deduce 
that 

F(k)^f'(y)ash^0 

where k = g(x + h) —g(x). 
For k^Ov/e have shown that 

f(g(x+h))-f(s(x)) g(x + h)-g(x) 
= F(k) . - 



h 



h 



But this equation is also true when k = since then both sides are zero. It 
follows that 



f(£(x + h))-f(g(x)) 



f'(y)g'(x) a sh^Q 



and this is what had to be proved. 



Differentiation 

10.14 Example D{\ + a-"} 100 = l°0O + * 97 )".97x 96 . 

Proof Set/0) =y 100 andg(x) = l+x 97 . By theorem 10.13, 
D{\ +x 97 } 100 - DfXgfx)) = f'(g(x))g'(x) 
= 100fe(x))".97x 96 
= 100{1 +x 97 }".97x 96 



99 



10.15 Exercise 

(1) If* >0 and «SN, prove that 

D{x vn \ = -x lln x-K 
n 

[Hint: exercise 3.1 1(6). Recall that the continuity of the nth root func- 
tion was considered in exercise 8.15(5).] 

(2) Use the previous question and the rule for differentiating a composite 
function (theorem 10.13) to show that, for any rational number r, 

D{x r ) = rx r ~' 

provided thatx>0. 

(3) Evaluate the following derivatives for those values of x for which they 
exist. 

(i) D{ 1 + x V2 Y s (ii) Dy/(X + V(* + V*)). 

' (4) Suppose that 

whenx=l. Prove that/'(l) = or /(l) = 1. 
(5) Let/: R->R be differentiable at x. Suppose that/admits an inverse 
function/" 1 : U-*U which is differentiable atj =f(x). Prove that 



DT l (y) = 



l 



Df(x) ■ 

[Hint: f' 1 °f(x) = x\ . The function g: R^-R defined by g(x) = x 3 
admits an inverse function #""': U-*U. Isg' 1 differentiable at the point 
0? 
t(6) A function/: R^-R is defined by 



m = 



x (x rational) 
— x (x irrational). 



Show that/o/(x) = x for all values of x. What may be deduced from 
theorem 10.13? 



11 



MEAN VALUE THEOREMS 



11.1 Local maxima and minima 

Let/be defined on an open interval (a, b) and let £ E {a, b). We say 
that /has a local maximum at £ if 

m<m 

for all values of x in some open interval / which contains %. Similarly for a local 
minimum. 





/ has beat max al ?. /has local min al {. 

Very roughly, /has a local maximum at £ if its graph has a 'little hill' above the 
point £. Similarly,/ has a local minimum at | if its graph has a 'little valley' above 
the point %. 

J f/(?) is the maximum value of/ on the whole interval (a, b), then obviously 
/has a local maximum at £. But the converse need not be true (see the diagrams 
above). 

11.2 Theorem Suppose that/is differentiable on (a, ft) and that £G (a, b). 
If /has a local maximum or minimum at |, then 

/'a) = o. 



100 





heal min 



Mean value theorems 



101 



Proof By definition 
/(*)-/(£) 



*-? 



/'(S)asx^?. 



Suppose that /'(£) > 0. From Exercise 8.15(6), it follows that for some open 
interval /=(£-/?,? + h) 



x-% 



>0 



(1) 



provided that xG 1 and xl=%. 

Let x, be any number in the interval (£ — h, %). Then x t — % < and hence it 
follows from (1) that/(x,) < /(£). Thus /cannot have a local minimum at £. Let 
x 2 be any number in the interval (£, £ + h). Then x 2 — % > and so it follows 
from (1) that/(x 2 ) >/(!). Thus/cannot have a local maximum at %. 




/'(£)> 



{ *i 'b 



A similar argument applied to —/deals with the case when /'(?) < 0. The only 
remaining possibility is/ (if) = 0. 



1 1 .3 Stationary points 

A point | at which /'(£) = is called a stationary point of/. Not all 
stationary points give rise to a local maximum or local minimum. The reader 
will be familiar with the case when /'(?) = and /has a point of inflexion at % as 
in the diagrams below. 




f\i) = o 



102 



Mean value theorems 



But worse behaviour than this is possible. The diagram below illustrates a case 
where /'(?) = but /has no local maximum, no local minimum, and no point of 
inflexion at £. (See the solution to exercise 16.5(6).) 




11.4 Theorem (Rolle's theorem) Suppose that /is continuous on [a, b] and 
differentiable on (a, b). If /(a) = f(b), then, for some % G (a, b), 

f'Q) = 0. 



/'(?) = 



Xa)=f(b) — 




Proof Since /is continuous on the compact interval [a, b],it follows 
from the continuity property (theorem 9.12) that/attains a maximum M at 
some point £i in the interval [a, b] and attains a minimum m at some point g 2 in 
the interval [a, b]. 

Suppose £, and % 2 are both endpoints of [a, b] . Because f(a) = f(b) it then 
follows that m = M and hence / is constant on [a, b] . But then /'({•) = for all 

Suppose that | t is not an endpoint of [a, b] . Then £, G (a, b) and /has a local 
maximum at £,. Thus, by theorem 11. 4, /'(£]) = 0. Similarly if | 2 is not an end- 
point of [a,b]. 
[Note the importance of the continuity property in this proof.] 



1 1 .5 Mean value theorem 

U-6 Theorem {mean value theorem) Suppose that /is continuous on [a, b] 
and differentiable on (a, b). Then, for some % G (a, b), 



Mean value theorems 

Kb) -m 



103 



/'(«) = 



b-a 



In the diagram, the slope of the 
chord PQ is 

f(b)-Ka) 
b-a 

Thus, for some % G (a, b), the tangent 
to /at % is parallel to PQ. 




Proof LetFbe defined on [a, b] by 

F(x) = f(x) + hx 

where h is a constant. Then F is continuous on [a, b] and differentiable on 
(a, b). We choose the constant /( so that F(a) = F(b). Then 



and so 



/(«) + ha = f(b) + hb 



h = - 



m -m 



b-a 



Since F satisfies the conditions of Rolle's theorem, F '(£) = for some 
% G (a, b). But then 

F'{%) = f'm + h - 
and the theorem follows. 

11.7 Theorem Suppose that /is continuous on [a,b] and differentiable on 
(a, b). lff'(x) = for each x G (a, b), then /is constant on [a, b] . 

Proof Let y G (a, b] . Then /satisfies the conditions of the mean value 
theorem on [a,y] . Hence, for some £ G (a,y), 



f'CO = 



m-m 

y-a 



But /'(£) = and so f(y) = /(a) for any y G [a, b\ . 

(Note: This theorem will be invaluable in chapter 13, about integration. It may 
seem intuitively obvious but, without the mean value theorem (which depends 
ultimately on the continuity property), it would be very hard to prove.) 



11.8 Exercise 

(1) Find the stationary points of the function /: U ~* D 



defined by 



104 Mean value theorems 

f(x) = x(x — l)(x — 2). Determine the maximum and minimum values 
of/on the interval [0, 3] . 

(2) Suppose that /and g are continuous on [a, b] , differentiable on (a, b) 
and that g'(x) =£ for any x G (a, b). Prove that, for some | S (a, b), 

g'm g{b)-g{a)- 

This is the Cauchy mean value theorem. [Hint: consider F = f+ hg 
where h is a constant.) 

(3) In addition to the hypotheses of question 2, suppose that /(a) = 
g(a) — 0. Deduce that 

b « - to as 

provided that the second limit exists. This is L'Hopital's rule. [Hint: 
exercise 8.15(4).] 

Use L'Hopital's rule to evaluate 

lira {y-y/V+y 1 )} 



is differentiable at every point and that 



[Hint: put_y = x l . 
(4) Suppose that /: U - 

/'(*) = x 2 

for all x. Prove that f(x) = |ar 4- c, where c is a constant. 
* (5) Let /: U •* U be n times differentiable at every point. Let P be a 
polynomial of degree n — 1 for which 

mi) = pai) CF-o. !,...«) 

where | > £i> ■ ■ ■ , % n are a 'l distinct points. Prove that, for some |, 
/ <n) (?) = 0. 

*(6) Letg-: U~*-R satisfy ^(0) =^(1) = and suppose that/: R->R is 
differentiable at every point and y = f(x) is a solution of the differen- 
tial equation 

dy 
g(x)f+y = 1. 
dx 

Prove that/(x) = 1 for any x satisfying < x < 1 . [Hint: Use Rolle's 
theorem to show that any open subinterval of (0, 1) contains a point £ 
at which /(|) = 1 . Then appeal to the continuity of /in the manner of 
exercise 9.17(2).] 



Mean value theorems 
1 1 .9 Taylor's theorem 



Suppose that /is 
differentiable at the point £. Then 
the equation of the tangent to 
y = fix) at the point % is 

y =/© + (* -£)/'«)• 



105 



y=f(x) 




s'y = M) + U- ?)/'«) 






Thus, for values of x which are 'close' to £, it is reasonable to expect that 
/(£) "•"(•*"" £)/'(£) i s a 'good' approximation to/(x). But how large may the 
error in this approximation be? 

Suppose now that /is n — 1 times differentiable at the point %. Then the first 
n — 1 derivatives at the point % of the polynomial 

p(x) = /(?) + yr(* - tr'cu + ^ - ?) 2 /"od + • • • 
+ ^ L T)i (x " r " /< "" 1>(l) 

are the same as those of/. (See exercise 10.1 1(5).) 



We might therefore reasonably expect that 
P(x) will be a 'very good' approximation to 
f(x) for values of x which are 'close' to £. But 
again, how large may the error be in this 
approximation? 



,->' 








11 

1/ y 


= /'(.v) 


y = fix) V 

— — — — ~y 






' 




>- 



i 



These questions are answered to some extent by the following version of 
Taylor's theorem. If n = 1, the theorem reduces to theorem 1 1 .6 (the mean 
value theorem). One may therefore regard Taylor's theorem as the Kth order 
mean value theorem. 

11.10 Theorem (Taylor's theorem) Suppose that /is n times differentiable on 
an open interval / which contains the point %. Given any x G I, 

fix) = m) + ^ ( X - !)/*<D + i (x - oY'm + ■■■ 



+ 



i 

(«-l)! 



&-& , - 1 f i *-%) + E„ 



106 Mean value theorems 
where the error B„ satisfies 

n\ 

for some value of 7? between if andx. 

Proof liy lies between x and g, put 

^(y) = fix) -f(y) - (x -^)/'(y) - ... - ( *~_^" ) , 1 / ' ' "CO- 
Taylor's theorem asserts that, for some t? between .* and g, 

m) = ^(x-m in \n) 

n\ 
The function F has the simplifying property that 



Mean value theorems 



107 



(2) 



F'O) = 
The function 



(» - 0! 



/<»>(y) 



has the property that G(£) = G(x) = 0. It follows from Rolle's theorem that 
there exists an 77 between x and % for which 

= G'cn) = nq)+ irtr *TS) 

(n-l)! 7 w ((x-|) n rw 
Identity (2) follows. 



but/ (n) (g) # 0. In each of the following cases decide whether or not / 

has a local maximum or a local minimum at g. In those cases where/ 

has neither a local maximum nor a local minimum discuss its behaviour 

close to the point £. 

(i) / <n> (g) > and n even. 

(ii) / ( "%)>0andHodd. 

(iii)/ <n) (S)<0 and n even. 

(iv)/ (n) (£)<0and«odd. 

[Hint: exercise 8.15(6).] 

Discuss the behaviour of the functions given below at the point 0. 



(v)/(x) = x 93 (vi)g(x) = x 94 (vu)A(x) = -x 



96 



72.77 Exercise 

(1) IfnENandxE R, use Taylor's theorem to prove the binomial 
theorem in the form 



n(n-\) , 
(1 + *)" = 1 +nx+ K x 2 



x . 



(2) Use Taylor's theorem with n = 3 and £ = 4 to obtain an approximation 
to y/5 for which the error is at most 2" 9 . 

(3) Let /be n times differentiable on an open interval 7 which contains the 
point if and suppose that D"/is continuous at the point g. You are 
given 

fm = /"© =...=/ <n " l) (£) = 



MONOTONE FUNCTIONS 



12.1 Definitions 

Let /be defined on a set S. We say that f increases on the set S if and 
only if, for each x G S and y G S with x <y, it is true that 

If strict inequality always holds, we say that /is strictly increasing on the set S. 

Similar definitions hold for decreasing and strictly decreasing. 

A function which is either increasing or decreasing is called monotone. A 
function which is both increasing and decreasing must be a constant. 



12.2 Examples The function/: K -*U defined by f(x) =x 3 is strictly 
increasing on R . The function/: (0, °°) ->R defined by f(x) = l/x is strictly 
decreasing on (0, °°). 

y 





12.3 Limits of monotone functions 

We introduce the notation 

/($-) = lim f{x); /(?+) - lim /(*) 
x->$- x-*i+ 

provided that these limits exist. Do not confuse /(£ — ) and /(£ +) with/(£), 
which is the value of the function/at £f. In example 8.5, /(l — ) = = f(\)but 
/(l +) = 2. 

108 



Monotone functions 



109 



12.4 Theorem 

(i) Let /be increasing and bounded above on (a, b) with smallest upper 
bound L. Then f{x) -* L as x -* b — . 

(ii) Let /be increasing and bounded below on (a, b) with largest lower bound 
/. Then f{x) -> / as x -*a +. 

Proof We only prove (i). Let e > be given. We have to find a 5 > 
such that, given any x satisfying b — 8 < x < b. 



i.e. 



l/<X>-/.Ke 

L-e<f(x)<L + e. 

The inequality f(x) < L + 6 is automatically satisfied because L is an upper 
bound for/on (a, b). Since L - e is not an upper bound for/on (a, b), there 
exists a y G (a, b) such that /( y) >L-e. But /increases on (a, b). 
Therefore, for any x satisfying 
y<x<b, 

L-e<f(y)<f(x). 

The choice 5 = b —y then completes 
the proof. 




12.5 Corollary Let /be increasing on (a, b). If £ G (a, b), then /(£ — ) and 
/(£ +) both exist and 

/(*) </(i -) </(?) </(? +) </(y) 

provided that a <x<%<y<b. 



/(£+> 




Ptoo/ The function /is bounded above on the interval (a, %) by /(£). 
By theorem 12.4, the smallest upper bound is/(£ -). It follows that, for any 



110 Monotone functions 

fix) </(*-) </(*). 

A similar argument for the interval (£, 6) yields the other inequalities. 



12.6 Differentiable monotone functions 

12. 7 Theorem Suppose thai /is continuous on fa, b] and differentiable on 
(a, b). 

(i) If f'{x) > for each x G (a, b), then /is increasing on [a, b] . If fix) > 
for each x G (a, b), then /is strictly increasing on [a, b] . 

(ii) If f'(x) < for each x G (a, b), then /is decreasing on [a,b]. If f\x) < 
for each xE(a,b), then/is strictly decreasing on [a,b]. 

Proof Let c and d be a/ty numbers in [a, b] which satisfy c <d. Then 
/satisfies the conditions of the mean value theorem on [c,d\ and hence, for 
some if G(c,rf)> 



fm = 



f(d)-m 

d-c ' 



If /'(*) > for each x £ (a, b), then /'(£) > and hence f(d) >f(c). Thus / 
increases on [a, b]. 

If fix) > for each x G (a, b), then /'(£) > and hence fid) >/(c). Thus / 
is strictly increasing on [a, b]. 

Similarly, for the other cases. 



12.8 Example Consider the function/: R-*-|R defined by f(x) =x(l —x). 
We have f\x) =\-2x. Hence f(x) > when x < \ and /'(*) < when x > \ 
It follows that / increases on (— °°, j] and decreases on \\, <»). 




12.9 Inverse functions 

A strictly increasing function /defines a 1 :1 correspondence between 



Monotone functions 



111 



its domain / and its range /. Hence / always has an inverse function / "' and this 
is strictly increasing on J. (Proof?) 





/-■?/— / 



In the case when / is an interval and / is continuous on /, then J = /(/) is also 
an interval (theorem 9 .9) . Observe that / _1 : J -* / is continuous on /. This is a 
simple consequence of corollary 12.5. (If/" 1 had a discontinuity at % G/, then 
we could find a X G / for which /(X) would be undefined.) 



/ ' discontinuous at if 

r 1 (£-)<x </-'(£+) 

/(X) undefined 




The next theorem is usually remembered in the form 

dx = (dy\ 
dy \dx) 

although this notation begs a lot of questions. 

12.10 Theorem Let / and / be intervals and let 7° and J° be the correspond- 
ing open intervals with the same endpoints. Suppose that /:/->/ is continuous 
on I and that /=/(/). 

If /is differentiable on 1° and 

£>/(*) >0 ixEI") 

then f' v :J^I exists and is continuous on /. Also / "' is differentiable on J" and 



112 Monotone functions 



Df~\y) = 



l 



(ye J") 



as/z->0. 



Df(x) 
provided that y = fix). 

Proof It follows from theorem 12.7 that /strictly increasing on /and 
hence/"' :/-*•/ exists. The continuity of/" 1 has already been discussed. There 
remains the consideration of its derivative. 

Let y G/°. Then x =/"'(j) S 1°. Put k =f~\y + h) -f~\y). Then 
/"'(;' + h) =f-'(y) + k=x + k. Thusj> + h = f(x + k) and it follows that 

h =f( x + k)-y =f(x + k)-f(x). 

Since/" 1 is continuous on J, k ■* as h ■* 0. Also,/" 1 is strictly increasing 
and so k & unless h =0. 

We may therefore appeal to theorem 8.17(ii) to obtain 

r l (.y + fQ-r l (y) m k | i 

h f{x + k)-f{_xff\x) 

Note that all the above results apply equally well to strictly decreasing func- 
tions. 



12.11 Roots 

As an application of our observations on inverse functions, we now dis- 
cuss the theory of nth roots. 

Let n be a natural number and consider the function /defined on [0, °°) by 

fix) = x". 

Since /(0) = and f(x) -* + oo as x -> + °°, it follows from the continuity 
property (theorem 9.9) that the range of /is also [0, «). 
Observe that 

/'(*) = nx"- l >0 (x>0). 

Thus/is strictly increasing on [0, °°) and, as in theorem 12.10, it has an inverse 
function/" 1 . We write 











Monotone functions 



113 



This argument shows that, given any y > there exists exactly one x > 
(namely x =y u ") such that.y = x". Our basic assumption about nth roots made 
in §1.9 has therefore been justified. 

In exercise 10.15(1), we calculated the derivative of the nth root function. 
We can therefore check the validity of the formula of theorem 12.10 in this case. 



^ 1/n = sr- = ^ 



— i — _ i y i/« y -i 

n{y Un f-' n y y ' 



If m is an arbitrary integer, the rule for differentiating a composite function 
(theorem 10.13) then yields 



D(y m ' n ) = D(y m ) iln = ^() 



Un y -m my m-l = ^^m/n^-1 



12.12 Exercise 

(1) Suppose that /increases on the compact interval [a, b] . Prove that / 
attains a maximum at b and a minimum at a. 

(2) Let n G N. Prove that the function /: [0, °°) ■* R defined by 

f( X ) = (x + l) 1 



\iln _ x l/n 



decreases on [0, °°). 

(3) Suppose that /is differentiable and increasing on an open interval /. 
Prove that f\x) > for each x G /. If /is strictly increasing on /, does 
it follow that /'(*) > for each x G /? Justify your answer. 

(4) Let /: R -* U be defined by f(x)= I + x + x 3 . Show that / has an 
inverse function/" 1 : R-*R . Calculate the value of £>/"■( y) when 
y=-\. 

T (5) A function / increases on the interval / and for each a and b in / it is 
true that, if X lies between f(a) and fib), then a % G / can be found 
such that /(?) = X (see corollary 9.10). Prove that /is continuous on /. 
Does the same conclusion hold if the hypothesis that /increases is 
abandoned? 

t(6) Let/: [0, 1] -*-(0, °°) be continuous on [0, l].LetM: [0, 1] ->(0,°°) 
be defined by 

M{x) = sup f(y) (0<x<l). 

^y <x 

Prove that the function 



<*>(x) = lim 



Mix) 



is continuous if and only if/ increases on [0, 1]. 



1 14 Mono tone functions 

1 2. 1 3 Convex functions 

Let /be defined on an interval /. We say that /is convex on / if it is 
true that, for each a > and (3 > with a + (3 = 1 , 



1 



f(ax+py)<af(x)+0f(y) 



(1) 



whenever x G/andy E.I. If the inequality sign is reversed, we say that /is con- 
cave on /. It is obvious that /is concave on /if and only if —/is convex on /. 

The geometric interpretation is that any point on a chord drawn to the graph 
of a convex function lies on or above the graph. 



a/(.v)+i3/0-) 

/(O.Y+0V) 





/convex 



/ concave 



It is sometimes useful to rewrite the definition of a convex function in the 
following way. A function /is convex on an interval / if and only if, for all 
pointsx,,x 2 and x 3 on the interval / satisfying x, <x 2 <x 3 , 

/(*i)-/(*i) ^ /(* 3 )-/(* 2 ) 

Equivalently, a function /is convex on an interval / if and only if for all 
points x lt x 2 andx 3 in the interval satisfying Xj<x 2 <x 3 , 

f(X2)-f(x 1 ) < f(x 3 )-f(x l ) 



(2) 



Xo-X, 



x*—x. 



(3) 




-v 3 .v, 
(2) slope P,P,< slope P,P 3 



(3) slope P,P, < slope P,P S 



That (2) and (3) have the same content as (1) is easily seen by making the 
substitutions x, =x,x 2 = ax + (Sy and x 3 =y (see exercise 1 2.21(2)). 



Monotone functions 
12.14 Tlieorem Suppose that /is convex on the open interval /. Then the 

h _>o- h >>-»o+ « 

both exist for each x£/. 

Proof Suppose that < ft, < ft 2 . Put x = x,, x 2 = x + ft, and 
;c 3 = x + ft 2 in (3). Then 

/(x + ft,) -/(*) /(x + ft 2 )-/(x) 



ft 2 



Hence the function 



F(ft) = 



/(x + ft)-/(x) 






increases in some interval (0, 6). The existence of 
lim FQi) 



h-*a+ 



then foUows from theorem 12.4. A similar argument establishes the existence of 
the left hand limit. 



12.15 Example The function/: U^U defined by /(x) = \x I is convex on 
R . (Tlie chords lie above the graph). 
Note however that the two limits of 
(4) are not equal when x = 0. 



J- = 1x1 




12.16 Theorem Suppose that /is convex on the open interval /. Then /is 
continuous on /. 

Proof From theorem 12.14, 

( , f(x + ft )-/00 

lim {/(* + ft) -/(*)} = hm 7— 

h->o- '•— ■"- 



h-*0- 



lim ft) =0. 

h->-0- 



Similarly, 



lim {/(x + ft)-/(x)} - 

h-*0 + 



,. f(x + h)-f(x ) 
lim : 



h-»0 + 



lim ft| = 0. 

h-*0+ 



116 Monotone functions 

12.17 Example The function/: R-> R defined by 
1 (|x|<I) 



/(*) = 



2 (|x|>l) 



is convex on the compact interval 
[—1,1]. Observe that it is not 
continuous on [— 1 , 1 ) . 



E%%% <//////& 



Example 12.15 shows that a function /which is convex on an open interval / 
need not be differentiable on /. However, iff is differentiable on /, then the 
following theorem provides a useful characterisation of convexity. 

12.18 Theorem Let / be an open interval and suppose that /is differentiable 
on /. Then /is convex on /if and only if Df increases on /. 

Proof (i) Suppose first that Df increases on /. Let x, <x 2 <x 3 . By the 
mean value theorem 

- /(£)> - = / (t?) 



X 2 ~X, 



X-i—X-, 



where x, < % <x 2 <r\<x 3 . Since Df increases, /'(?) </'(t?) and therefore 
inequality (2) holds. Thus /is convex on /. 

(ii) Suppose next that / is convex on /. Let x , < x 2 < x 3 < x 4 . By inequality 

/(*a)~/(*i) < /(* 3 ) -f(x 2 ) /(x 4 )-/(x 3 ) 



x-> — x, 



X,—Xi 



Xa 



Ignore the middle term in this inequality and let x 2 -> x , + and x 3 -> x 4 — . We 
obtain /'(x,) </'(* 4 ) and it follows that Df increases on /. 

If /is twice differentiable on /, an even simpler characterisation of convexity 
results. 

12.19 Theorem Let / be an open interval and suppose that /is twice diffe- 
rentiable on /. Then / is convex on / if and only if 

f"(x)>0 
for all x e I. 

Proof The theorem follows from theorems 12.7 and 12.18. 



Monotone functions 



117 



12.20 Examples The functions/,^ and h defined on (0, °°) by /(x) -x\ 
g(x) = 1/x and h(x) = x+ x~ l ate all convex on (0, °°) because their second 
derivatives are all non-negative on (0, °°). 

The function 4> defined on (0, °°) by 4>(x) = s/x is concave on (0, »). We 

<p"(x) = -^' 3/2 <0 (*>0). 



12.21 Exercise 

(1 ) Let /: R -* R be defined by f(x) = (x - l)(x - 2)(x - 3). Find the 
ranges of values of x for which /is (a) increasing, (b) decreasing, (c) 
convex, (d) concave. Draw a graph. 

(2) Let /be convex on the interval / and let x x , x 2 and x 3 be points of / 
which satisfy x, <x 2 <x 3 . Prove that 

/(*a)-/(*i ) < /(*3)-/(*i) < fix 3 )-f(x 2 ) 



x 2 -x x 



X 3 -Xi 



x 3 -x 2 



(3) Let /be convex on the open interval /and let J = /(/). If /is strictly 
increasing on /, show that/" 1 is concave on J. What happens if /is 
strictly decreasing on 11 
*(4) Suppose that /is convex and differentiable on the open interval /. If 
% G /, prove that 

f(x)-m)>f'(%\x-%) (xe/). 

Interpret this result geometrically. What result is obtained if /is con- 
cave on/? 

Let 0: U -*■ R be a strictly increasing, convex, differentiable function 
on R and suppose that 0(£) = 0. If x, > % and 



= x„ — 



0(*n) 



(a = 1,2,...) 



prove that x„ •* $. as n ■* so. (This method of obtaining an approxima- 
tion to the zero of is called the Newton-Raphson process.) 

+ (5) Let /be differentiable, convex and bounded on R. Show that /is a 
constant. 

"I" (6) Let /be continuous on an interval /and satisfy 



/ 



'x+y 



, m+f(y ) 

2 



for each x E I and y E I. Prove that, for any x , , x 2 , 
interval /, 



, x„ in the 



/(— "I < ~ ifiXx)+f{X2) +■■■ +«*»)>. 



118 Monotone functions 

[Hint: recall the induction argument of example 3.10.] Deduce that/ 
is convex on /. [Hint: establish inequality (1) of §12.13. Begin with the 
case when a and are rational.] 



1 






13 



INTEGRATION 



13.1 Area 



Suppose that /is continuous on the compact interval [a, b] . 



Does it make sense to discuss the 'area' under 
the graph of/? And, if so, how can we com- 
pute its value? 



If it makes sense at all to talk about the area under the graph, then presum- 
ably this area must be at least as big as the areas S] and S 2 which have been 
shaded in the diagrams below. 






Let S denote the set of all numbers S which can be obtained as the sums of 
the areas of little rectangles as in the diagrams above. Then the 'area' under the 
graph of /must be at least as big as every element of S. It seems reasonable to 
identify the 'area' under the graph of /with the smallest number larger than 
every element of S, i.e. with sup S. 

These remarks are intended to motivate the formal mathematical definition 
given in the next section. 



119 



120 



Integration 



13.2 The integral 

Suppose that /is continuous on the compact interval [a, h]. We pro- 
pose to give a definition of the integral 



f> 



dx 



of /over the interval [a, b]. 

We begin by defining a partition P of the interval [a, b] . This is a finite set 
P= [vo, y\, ■ ■ ■ ,y n ] of real numbers with the property that 

" = yo<yi<yi< ■ ■ -<y n -i<y n = b. 

Since /is continuous on [a, 6], it is bounded on [a, b].It therefore makes 
sense, given a partition P = {y Q ,y u . . . ,y n } of [a, b],to define 

m, = inf. f(x) 

y„«S x < y, 
™ 2 = ^"t ft*) 

and so on . 

For each partition P, we can then form the sum 



S(P) = E^feOft-^-i). 



k = i 



The diagram below illustrates the case n = 4. The value of S(P) is the area of 
the shaded region. 




>'i >'2 j'j y* 



Suppose that / is bounded above on [a, b] by H. Thus /(/) <Hfor any 
r S fa, ft] . Then 

S(f) = E WfcO'fc-J'fc-i) 



<^ ZOfe-^-i) = -W{Oi-^o) + ... + 0' n -/„-:)} 



fe = i 



= H(y n -y ) = H(b-a). 



Integration 



121 



It follows that the set of all numbers S(P), where P is a partition of [a, b\ , is 
bounded above by H(b —a). Hence it has a smallest upper bound and we may 
define 

f f(x)dx = sup5(P) (1) 

Ja p 

where the supremum extends over all partitions P of [a, b] . 
We wish to think of 



J>> 



dx 



as the 'area under the graph off. But notice that any area below the x-axis has 
to be counted as negative for this interpretation. 
Note also that, if b > a, we define 



C f(x)dx = -f*f(x)dx. 



1 3 .3 Some properties of the integral 

Much of the power of the idea of integration lies in its connexion with 
differentiation. For example, it would be very painful indeed if all integrals had 
to be worked out directly from the definition. Our first priority is therefore to 
prove the theorems which link integration with differentiation. Before we can 
do this, however, we need to derive a few basic results directly from the 
definition. 



13.4 Proposition Let /be continuous on [a, b] with maximum M and 
minimum m. Then 



m 



(b-a)<{ f(x) dx<M(b-a). 



The proof of this proposition is essentially contained in §13.2. We quote two 
simple corollaries. 

13.5 Corollary If c is a constant, then 

"b 



cdx = c(b—a). 



13.6 Corollary Let /be continuous on [a, b] and satisfy |/(r)l < k for any 
t £ [a, b] . Then, for any £ and x in the interval [a, b] , 



JlfiOdt 



<k|x-|| (x*&. 



122 Integration 

The next proposition is also very simple, provided that one recalls the proper- 
ties of sup and inf. (See exercise 2.13(2) and exercise 7.16(6i).) 

13. 7 Proposition Let /be continuous on [a, b\ and let c be a constant. Then 
J„ {f(t) + c}dt = \ b a f(t)dt + c(b-a). 

The next and final result of this section is not quite so easy and so we give the 
proof. 

13.8 Theorem Let /be continuous on [a, b] and let a<c< b. Then 



j a f(t) dt = £ f(f) dt + j fit) dt. 



Proof We use the notation of § 13.2. Suppose, in the first place, that 
Pi and Pi are any partitions of [a, c] and [c, b] respectively. Then the set P of 
points in at least one of the sets P, and P 2 is a partition of [a, b] . 

„D " i i i i i i ii i L 



p, 



/>, 



It is obvious that S(P) = S(P t ) + S(f> 2 ). Moreover, 

~b 



S(P)<j a f(t)dt 

because of definition (1) of §13.2. Thus, given any partition P x of [a, c] and any 
partition/^ of [c,b], 

siPi) + s(p 2 ) <[/•«) dt 

S(Pd<j*f(t)dt-S(P 2 ). (2) 

Hence, given any partition P 2 of [c, b], the right hand side of (2) is an upper 
bound for the set of all numbers of the form S(P{) where />, is a partition of 
[a, c] . Therefore 



sup5(P,)< f f(t)dt-S(P 2 ) 



where the supremum extends over all partitions P t of [a, c] . Recalling the 
definition of an integral, we obtain 






and so 



Integration 

£ fit)dt< j b a f(t)dt-S(P 2 ) 

s(p*)<j b a mdt-\ L a f(t)dt. 

A similar argument now yields 

J" c " fit) dt < £" fit) dt - jl fit) dt 



123 



and therefore 

•b 



j b a f(t)dt>f a f(t)dt + j € f(t)dt. 



(3) 



Now let P be any partition of [a, b] and let Q be the partition obtained from 
P by inserting the extra point c (if it does not already belong to P). It is easily 
seen that 

S{P) < 5(0. 

Let Pi be the partition of [a, b] consisting of those points of Q which lie in 
[a, c] and let P 2 be the partition of [c, b] consisting of those points of Q which 
lie in [c,b]. 



a [ II HI I 11 1 I I I I ],, 



We have 



SiP)<S(Q) =SiPd+Sm 

<j° a fit)dt + j* fit)dt. 



The right hand side of this inequality is therefore an upper bound for the set of 
all numbers S(P) where P is a partition of [a, b] . Hence 



sup S(P) < f a f(t) dt + j c f(t) dt 



i .e . j a fit) dt < j C a fit) dt + £ fit) dt. (4) 

Combining (3) and (4), we obtain the conclusion of the theorem. 



124 Integration 

13.9 Differentiation and integration 

Let /be defined on (a, b). Suppose that F is continuous on [a, b] and 
differentiable on (a, b) and satisfies 

F\x) = fix) 

for each x G (a, b). Then we shall say that F is a primitive (or anti-derivative) of 
/on [a, 6]. 



13. 1 Example lff(x) = x 3 , then the function F defined by F(x) = Jx" + 3 
is a primitive for/ 



75.77 Theorem Let fbe a primitive for/on [a, b] and let G be defined on 
[a, b] . Then G is a primitive for/on [a, b] if and only if, for some constant c, 

G(x) = F(x) + c 

for each x G [a, b] . 

Proof Obviously F(x) + c is a primitive for /on [a, 6] . Suppose there- 
fore that G is a primitive for /on [a, b] . Then F-G is continuous on [a, b], 
differentiable on (a, b) and, for each x G (a, b), 

D{F(x)-G(x)} = F'(x)-G'(x) = /(*)-/(*) = 0. 

From theorem 1 1 .7 it follows that F — G is constant on [a, b] . Hence the 
result. 

What have primitives to do with integration? Suppose that/is continuous on 
[a, b] . Then we may define a function Fon fa, b] by 

%) - J* f®4t (a<x<b). 

In the next theorem, we shall show that F is a primitive of/ on [a, 6] . But 
first we give a rough, intuitive argument to indicate why one might suppose this 
to be true in the first place. Given x G (a, b), we seek to explain why one might 
suppose that F'(x) -f(x). 

The number F(x) may be interpreted geometrically as the 'area under the 
graph of /between a and x'. The shaded region in the diagram below should 
therefore have area F(x + h)—F(x). 



/•(.v + h)-F{x) 




x x + h ft 



Integration 



125 



Since /is continuous, one feels that, if h is very small, the shaded region will 
'nearly' be a rectangle and hence its area will be approximately hf{x). Thus 



and 



F(x + h)-F(x) = hf(x) 



F(x+h)-F(x) . 



= fix). 



(5) 



This equation is only approximate, but the smaller h becomes the better the 
approximation should be. Allowing h -* 0, we might therefore expect that 
F'(x)=f(x). 

This argument is dubious on a number of counts - particularly at the stage 
where we divide by the small number h to obtain (5). In the next theorem we 
provide a precise proof of the result. 



75.72 Tlieorem Suppose that /is continuous on [a, b] and that F is defined 
on [a, b] by 



F(x) m f* f{t) dt {a<x< b). 



Then F is a primitive of /on [a, b], i.e. F is continuous on [a, b], differenti- 
able on (a, b) and F\x) = f(x) for each x G (a, b). 

Proof Since /is continuous on [a, b] , it is bounded on [a, b] . Suppose 

that \f{t)\<K (a<t<b). 

By theorem 1 3 .8, for any x and % in [a, b] , 



F(x)-F® = }*/(') 



dt 



and therefore, by corollary 13.6, 

\F(x)-F(£)\<k\x-$\. 

And, from this inequality, it follows that F is continuous on [a, b] . 

A more subtle argument is needed to show that F is differentiable on (a, b). 
If x and | are in (a, b) and x ¥= %, then 

The last step is justified by proposition 13.7 withe =/(£). 



126 Integration 

Let e>0 be given. If £G (a, b), then /is continuous at the point % and so, for 
some 5 > 0, 

1/(0-/(91 <« 

provided that \t — £| < 5. Hence, provided that \x — %\ < 5, it follows that 
1/(0 —/(I) I < e for every value of t in the compact interval with endpoints £ 
and*. From corollary 13.6, we conclude that 



F(x)-F® 



x-% 



-/(I) 



1 



< 



I* -SI 

1 



.elx-il 



= e 

provided that < \x — £|< 5. But this is what is meant by the assertion 
F(x)-FQ) 



*-i 



•/($)as*-*{. 



We have therefore shown that F is differentiable on (a, b) and F'(|) = /(£) for 
each | £ (a, 6). 



75. 75 Example 



D 



c 



dt 



l+t 



100 



1 +x 



100 ' 



i5.i4 Theorem Any function /which is continuous on [a, b] has a primitive 
on [a, b] . If G is any primitive of/, then 

|* /MA = G(b)-G(a) = [G(t)] b a . 

Proof Let Fbe defined on [a, b] by 

F(x) = f /(r) dt (a < a: < 6). 

Then 

J a /(/)A = F(b) = F(b)-F(a). 

By theorem 13.12, Fis a primitive of/ on [a, 6] and, by theorem 13.11, any 
other primitive G satisfies 

G(x) = F(x) + c (a<x<b) 



Integration 



127 



for some constant c. Hence 

•b 



J /(f) dt = {F(b) + c}- [F(fl) + c} 
= G(b)-G{a). 



13.15 Example Since 
1 



D - 



2(1 + ; 2 ) 



(1 + r 2 ) 2 ' 



it follows from theorem 13.14 that 



i; 



o.+ff 



:dt = 



1 



2(1 + f) 



4 



4 



13.16 Riemann integral 

The integral 

•b 



/. /w* 



(6) 



was defined as the supremum of all areas like that shaded in the left hand dia- 
gram below. It would be equally sensible to define the integral as the infinum of 
all areas like that shaded in the right hand diagram below. 





If /is continuous on [a, b] , these two definitions yield the same result. The 
reason is simple. The use of the second definition would yield the result 

-j" -f(x)dx. 

But (6) and (7) are equal because F is a primitive for/ on [a, b] if and only if 
— F is a primitive for — /. 

If /is not continuous on [a, b] , the two definitions do not necessarily yield 
the same result. But, //they do yield the same result, we say that /is Riemann 



(7) 



128 Integration 

integrable and call the common number obtained from the two definitions the 
Riemann integral of /on [a, b] . 



13.17 Example We know that 

JJ X *dx = [W]h = i. 

To illustrate the idea discussed in § 13.16, we evaluate the integral without 
using the theory of differentiation. 





Introduce the partition P n = {0, \/n, 2/n, 3/n, . . . , 1} of [0, 1] . The shaded 
area of the left hand diagram is smaller than the integral and the shaded area of 
the right hand diagram is larger. Thus 

fit \ n ) n Jo tx \n] n 

^{0 2 +l 2 +... + («-l) 2 }<^ x 2 dx< L{ l 2 + 2 l +...+„*} 

- 3 .-(n-l)n(2n-l)< f ' x 2 dx <- 3 .-«(« + l)(2#i + 1) 
» o ju no 



Considering the limit as n •* °° yields 



a £fe<f ll+^l ll+ ! 



(exercise 3.1 1(1 i)) 



Integration 



129 



13.18 Exercise 

(1) Prove that, given any natural number «, 



J>" 



dx = 



1 



B-f 1 



Show that the same result holds if?? is replaced by any rational number 
a =£ — 1 provided that < a < b. 
(2) Since the integrand is a square, the following result must be wrong. 
Explain why. 

2 



f 2 ax 
Jo (x-l? 



1 



x-l 



= -1-1 = -2. 







(3) If a is a positive rational number, prove that 
Hffl -ir{l a + 2 a + 3 Q + ...+«"} = 

" - ■» « 



a+ 1 



13.19 More properties of the integral 

In § 13.3 we mentioned some properties of the integral. Theorem 13.8 
is particularly important but its proof which we obtained directly from the 
definition was long and laborious. We used the properties examined in § 13.3 to 
prove the theorems which connect integration with differentiation. Having 
proved these theorems, we can now obtain the rest of the properties of the 
integral very easily. 

13.20 Theorem Suppose that /and g are continuous on [a, b] and that X and 
H are any real numbers. Then 

\" {x/(o + ng(t)} dt = x \ b a f{t) dt + n j* m dt. 

Proof Let F and G be primitives of/ and g respectively on [a, b] . By 
theorem 10.9(i) the function H = XF + jdG is a primitive of X/+ fig on [a, b] . 
Hence, by theorem 13.14, 

f " {X/(0 + pgffi dt = [XF(t) + M G(r)] b 

= HF(t)]Z + n[G(t)] b a 

= \f*f(t)dt + vj b a g(t)dt. 

13.21 Theorem (integration by parts) Suppose that / and g are continuous on 
[a, b] and have primitives F and G respectively on [a, b] . Then 



130 Integration 

\" b f(t)G(,)dt = [F(r)G(r)]2- P F(t)g(t)dt. 

Ja Ja 

Proof By theorem 10.9(h), 
D(FG) = fG+Fg 
and thus fG is a primitive of /G + Fg on [a, b\ . Hence 

H {f(t)G(t) + F(t)g(t)}dl = [F(f)G'(/)]* 

and the theorem follows. 

As is clear from its proof, the formula for integrating by parts is just the 
integral analogue of the formula for differentiating a product. The next result is 
the integral analogue of the formula for differentiating a composite function, i.e. 
a 'function of a function'. It is the rule for changing the variable in an integral 
and is usually remembered in the form 

dt 

dt = — . du. 

du 

13.22 Theorem Suppose that </> has a derivative which is continuous on [a, b] 
and that /is continuous on an open interval which contains the image of [a, b] 
under 0. Then 

Proof Let 

From theorem 10.13 and theorem 13.12, 
d 



du 



F(0(«)) = F'(<p(u))<p'(u) = /(0( W ))0'( W ). 



It follows from theorem 13.14 that 

j a f(<p(u))<p'(u) du = [F(0( W ))]£ 

and this is what had to be proved. 

13.23 Theorem Suppose that /and g are continuous on [a, b] and that, for 
any te [a, b],f(t)< g(t). Then 



( b f(t)dt<fg(t)dt. 

J a Ja 



Integration 



131 



Proof Let H be a primitive of g — /on [a, b) . Then 
DH(t) - g(t) —f{t) > (a < t < b). By theorem 1 2.1,11 therefore increases on 
[a, b] . Thus Hib) > //(a) and hence 

f {git)- fit)} dt = Hih)-Hia)>0. 

J a 

13.24 Theorem Suppose that /is continuous on [a, b]. Then 
f b fit)dt<[* 1/(01 dr. 

Proof Since - |/(/)| < fit) < 1/(01 (a<t< b), it follows that 
- [* |/(0I dt < J"* fit) dt < J a 6 1/(01 dt 
and hence the result. (Note that |/| = {/ 2 }" 2 and is therefore continuous.) 

13.25 Theorem (Cauchy-Schwarz inequality) Suppose that / and g are con- 
tinuous on [a, b] . Then 

Proof This is much the same as that of example 1.11. For any x, 
0<J fl " {xfit)+git)fdt 

= x 2 f {fit)} 2 dt + 2x\" fit)g(t) dt + I*" {git)} 2 dt 

JO J Q JO 

= Ax 2 + 2Bx + C. 

Thus the quadratic equation Ax 2 +2Bx + C = cannot have two (distinct) 
roots. Hence 

B 2 <AC 

as required. 



13.26 Exercise 

(1) Suppose that g is continuous on [a, b] and thatg(0 >0(«<f <S). 
Prove that 

\\it)dt = 

if and only if git) = for each r £ [a, b] . 



132 Integration 

(2) Suppose that /is twice differentiable on fa, b] and that/" is continu- 
ous on [a, b] . Prove that 

j b a xf"(x)dx = {bf\b)-f{b)}-{af(a)-f{a)}. 

*(3) Let /be positive and continuous on [1 , °°). Suppose that 

Prove that /(.v) > § (x - 1 ) (x > 1 ) . [Hint : The integral 
ft {F(t)}- 1/2 F'(t) dt is relevant.] 
*(4) Suppose that /and g are continuous on [a, b] and that g(t) > 
(a<r< b). Prove that 

l a mg<?)dt ~r®£ g(t)dt 

for some £ G [a, b] . [Hint: continuity property.] What has this to do 
with theorem 1 1 .6? 
*(5) Suppose that g is continuous on [a, b] and that /is differentiable on 
{a, b] and its derivative is continuous on [a, b] with f'(t) > 
(a<t<b). Prove that 

/„ f(t)g(t) dt m f(fl) j a g(t) dt + f{b) J* g(t) dt 

for some £ G [a, b] . What happens if f'(t) <0(a<t<b)l [Hint: inte- 
grate by parts.] 
(6) By integrating many times by parts, show that the error term in 
Taylor's theorem (theorem 1 1 .10) can be expressed in the form 



*« - &=w£ fr-o*- 1 /^) 



dt 



provided that/ (,,) is continuous on /. Show how the form of the error 
term given in theorem 11.10 can be obtained from that above. 



13.27 Improper integrals 

We write down some sample definitions. Suppose that /is continuous 
on [0, °°). Then we define 

j; D °/w^ = x iimj o x /w^ 

provided that the limit exists. Similarly, if /is continuous on (a, b] , then we 
define 



Integration 



133 



T f(x)dx = lim \" f(x)dx 

J-i-a y-*a* Jy 

provided that the limit exists. Note that, if/is continuous everywhere, we define 

P~ f(x)dx = f° f(x)dx + P~ f{x)dx 

provided that the appropriate limits exist, i.e. we allow the upper and lower 
limits to recede to +°° and -°° respectively independently. We do not define 
the improper integral by 



lim f(x) dx. 

■ — > DO J—X 



13.28 Examples 



o>r 



dx 

„2 



rx dx 
lim I —s = hm 
x->~Ji x 1 x->°> 



x 



-a. 1 -? r 1 - 



Jy X 



dx 

2 



On the other hand, if 0<y< 1, 

— = l-»+co a sy->0 + 

x y y 
and hence the improper integral j — does not exist. 

(ii)f 1 ^= lim f*- lim[2x" 2 ]i= Hm (2-2V>') = 2. 



X 


1 
J? 


^777 



v 1 






f ' dx 

J V* 

,/ — ► 



1 



I 



(iii) Note finally the following example. Observe that / ~ x dx does not exist 
(why not?). Hence • fjT.I * <& cannot exist even though 

|**xdx = U 2 -^ 2 ^0asAT^ + °°. 



The integrals defined in this section are called improper because they are not 
given by the definition of § 13.2. It is perhaps pedantic to include the little 



134 Integration 

arrows in the notation for improper integrals and most authors would simply 
delete them. But for those who propose to study integration more deeply, it is 
probably better not to do without the little arrows since this will avoid con- 
fusion with the differing definitions of the powerful Lebesque theory of integra- 
tion. 



13.29 Proposition Suppose that is a function which is continuous and non- 
negative on the open interval /. Suppose also that /is continuous on /and satis- 
fies 

\m\<m (xei). 

If the improper integral of <p over the interval /exists, then so does that of/. 

This is, of course, an analogue of the comparison test (theorem 6.15), for 
series and may be proved in much the same way. If/ happens to be non-negative 
on / a simpler proof may be based on theorem 12.4 concerning the convergence 
of monotone functions. 



13.30 Example To prove the existence of the integral 
dx 



f-»oo 

Ji r 



+ x 2 



we simply have to observe that, for any x > 1 , 
dx 1 

2^32- 



1 +X' 

.-2 



Since / "*~ x 2 dx exists, the result then follows from proposition 13.29. 



13.31 Euler-Madaurin summation formula 

We discuss only a simplified version of this useful result. 

13.32 Theorem Let /be continuous, positive and decreasing on [1 , °°). Then 
the sequence <A„> defined by 



An - t f(k)-C f(x)dx 



is decreasing and bounded below by zero. Hence it converges. 
Proof Since /decreases, for any k~ 1, 2, . . . , 
•ft + i 



/(* + !)< J/ f(x)dx<f(k) 



Thus 



Integration 



135 



a„ +1 -a„ = zm-j** 1 fix)**)- £/<*>-JT *«* 

= f(n + 1) - J"" + ' fix) dx </(« 4- 1) -f(n + 1) = 
and so <A„) decreases. Also 

> £/(*)- I A*) = /(«)> o 
fe=i fc=i 

and so <A n > is bounded below. 

From theorem 13.32 it follows that, if /is positive, continuous and decreas- 
ing on [1 , °°), then the series and the improper integral 

f A«) and C~ ftx)dx 

n = l Jl 

either both converge or else both diverge. The theorem therefore provides a 
criterion for the convergence of a series or an improper integral and for this 
reason is often referred to as the integral test. 



13.33 Example Take fix) = x" 1 in theorem 13.32. We know from theorem 
6.5, that the series S" =1 n~ l diverges to +°°. From theorem 13.32, we conclude 
that 

r "dx 



n x 



■ + °° as n -* °°. 



In particular, the improper integral / ^°° x" 1 dx does not exist. 



13.34 Exercise 

(1) Discuss the existence of the following improper integrals. 



... r -i dx ,.., f-" x 2 dx 

W Jo V^7) (ll) Jo ^^* 



«£ 



V(l -x) 
h dx 



(1 -x} 



3/2 



(v) 



(1 +x 3 ) 2 
*i dx 



«r i 



1 + X + x 1 



+ x+x s 



dx 



o il-x) 



(vi) — 

J^o X 



[Hint: for (v) and (vi) recall example 13.33.] 
(2) Prove that, for 0< y <\, 



136 Integration 



r(i'2)+y 2x — \, 

— -dx = 0. 

J(l/2)-y x(l —X) 

Discuss the existence of the improper integral 

J-, o 



■*i 2x- 



■dx. 



(3) Let a be a rational number with a > 1 . In theorem 6.6 it was shown 
that the series 

oo 1 

converges. Base another proof of this result on theorem 13.32. 



14 



EXPONENTIAL AND LOGARITHM 



14.1 Logarithm 

We define the logarithm for each x > by 

r* dt 



f * dt 
logx=J i - 



It follows immediately from the definition that log 1 = and that, for each 
x>0, 



Also, 



Dlogx = -. 
x 



D 2 \ogx = =. 



We conclude that the logarithm is strictly increasing and concave on (0, °°). 
Further, in view of example 13.33 and exercise 13.34 (lvi), 



and 



log x -*■ + °° as x -*■ + °° 



log x •* — °° as x -*■ +. 

(For an alternative proof of these results, see exercise 14.3(1) below.) With this 
information the graphs = log x can easily be drawn. 




The number e is defined by the equation 



137 



138 Exponential and logarithm 



log e = 1 . 
The value of e is approximately 2-718 (see exercise 15.6(4)). 

14.2 Theorem Suppose that x > 0, y > and r is rational. Then 
(i) \ogxy = log* + log .y 
(ii) logOO = r]ogx. 

Proof (i) For a fixed value of y > 0, consider the function 



f(x) = lOgATJ-logJC. 



We have 

xy xxx 

for each* >0. Thus /is constant (theorem 11.7) and 

log xy—\ogx = c (x> 0). 
To obtain the value of c, put x — 1. Then 
c = log y -log 1 = logj\ 
(ii) Consider the function 

fix) = log(* r )-rlogJC. 
We have 



f\x) = - r .rx r -'-- 

X X 



0. 



Hence /is a constant and 

\ogix r )-r\ogx = c. 
To obtain the value of c, put x = 1 . Then 
c = log 1 —/-log 1 = 0. 

Note The logarithm we are discussing here is the 'natural logarithm'. 
This is sometimes stressed by writing 'In' rather than 'log'. In pure mathematics, 
little occasion arises for the use of 'logarithms to the base 10' as found in the 
familiar logarithmic tables. 

14.3 Exercise 

(1) Prove that log 2 > 0. By considering log 2" and log 2~", show that 
log x -* + °° as x -* + °° and that log x -*■ — °° as x -* +. 

(2) Prove that, for each y > 0, logy <.y — 1 . If x is a positive rational 
number and x > 1 , deduce that 



Exponential and logarithm 



139 



logx^y. 

Given a positive rational number r, prove that 
(i)x" r logx ->0 asx -> + <» 
iii)y r logy ->-0 as>> -»0 +. 
(These results are usually remembered by saying 'powers drown 
logarithms'.) [ Hint : take < s < r when proving (i).] 
"(3) Show that Fix) = x log x - x is a primitive for log x on (0, °°). Deduce 
that 



i 



log(l+*)dx = 2 log 2-1. 



Hence show that 

C W» 

1 (2»)! 

log-' 



n 



w! 



- log — as n -*■ °°. 
e 



[Hint: recall example 13.17.] 
"(4) Show that the equation log x-ax has solutions if and only if ae < 1 . 
Show that 

elogx<x (x>0) 

with equality if and only if x = e. 

Suppose that x x > e and x n+1 = e log x„ (« = 1 , 2, . . .). Prove that 
x„ -*■ e as « -» °°. 
*(5) Prove the existence of a real number y (Euler's constant) with the pro- 
perty that 

1 H 1 1- ... -I log n -» 7 as « ■* °°. 

[Hint: theorem 13.32.] Hence show that 

h » 2 3 4 

(6) Evaluate 

(i) Z) {(log x) s } (JC > and s rational) 
(ii)i){loglogx} (X>1). 
Hence show that the series 

a i 



converges if r > 1 and diverges if r < 1. (See theorems 6.5 and 6.6 for 
comparison.) [Hint: theorem 13.32 again.] 



140 
14.4 



Exponential and logarithm 



Exponential 

From theorem 12.10 we may deduce the existence of an inverse func- 
tion to the logarithm. We call this inverse function the exponential function and 
write 

y = exp* if and only if x = \ogy. 

Since the logarithm function has domain (0, °°) and range R, the exponential 
function has domain R and range (0, °°). 



y — exp .v 




Observe that the exponential function is strictly increasing on R and that 
exp x-t + coasx-^ + c* 1 
exp x -*■ as x •* — °°. 



Also 



D exp x = 



1 



1 



— y — exp*. 



D\ogy \/y 
Thus the exponential function is its own derivative. 



14.5 ■ Exercise 

(1) If x andjc are real numbers and r is rational, prove that 

(i)exp(x+,y) = (exp x)(exp y) 
(ii) exp (rx) = (exp x) r . 
[Hint: These may be deduced from theorem 14.2. Alternatively, they 
may be proved directly. In the case of (i), for example, by differentiat- 
ing 

exp (x + y)/(exp x) 

with respect to x.] 

(2) Prove that, for any rational number r, 

(i) x~ r exp x-^ + ooasx-*-!- 00 
(ii) x r exp x -» as x -* — °°. 
(Roughly speaking, 'exponentials drown powers'.) 

(3) Use L'Hopital's rule (exercise 1 1 .8(3)) to prove that 



Exponential and logarithm 

m ,. expx-1 . .... .. log(l +x) 

(i) lim — = 1 (ii) hm = 1 - 

x-*0 * x-»0 x 

(4) Suppose that h is positive and continuous on [0, °°) and that 
h {x)>H(x){x>Q\ where 



141 



H(x) = 1 + h{t)dt. 
Jo 

Prove that, for any x > 0, 
h(x)>expx. 

(5) Determine the range of values of x for which the function /: 
defined by 

/(*) = exp{18* 3 } 

is (a) convex, (b) concave. 

(6) If /is continuous and increasing on [0, °°), prove that 



[ n f{x)dx< £/«< ["* i f(x)dx. 
Jo k% J ' 



Deduce that n log n — n < log n\ < (n + 1) log (n + 1) - n and hence 
show that 



— < exp n < ; . 

n! 



n\ 



14.6 Powers 

If a > and r is rational, then we have defined the expression a r . We 
propose to extend this definition to include non-rational exponents. 
If a > and x is real, we define 

a x = exp (xloga). 

But does this definition agree with our old definition in the case when x is 
rational? It follows from theorem 14.2(h) that, if r is rational, then 

exp {/-log a} = exp {log (a r )} = a r . 

Recall that e is defined by log e = 1 . Equivalently , e = exp 1 . Observe that 

e x = exp {x log e} = exp x 

which explains the familiar notation for the exponential function. 



142 Exponential and logarithm 



14.7 Exercise 

(1) If a and b are positive, prove the following. 

(i) a x * y = a x a y (ii) (ab) x = a x b x 

(iii) a" = \ (iv) (a x ) y = a xy . 

a 

(2) Prove that the derivative of a x with respect to x is (log a)a x . 

(3) Write 



1 + -| =ex P 



n log | 1 + - 



and hence show that 
lim 



= e" 



(see example 4.19). [Hint: theorem 8.9 and exercise 14.5(3ii).] 
(4) Write 



« ,/n = exp 



l -\« % n 



and hence show that n l,n ■* 1 as n -* °° (see exercises 4.20(6) and 
5.7(1)). 
(5) Discuss the existence of the following improper integrals. 



ȣ 



dx 



J-+0 J-f-oo 1 



+ X 2 



;dx. 



[Hint: proposition 13.29.] 
' (6) Obtain all solutions of the functional equations 

(i) f{x +y) = f(x) +f(y) (X 6 U,y GR) 

(ii) / (x +y)= fix) f(y) ( X € R,y S R ) 

(iii) /(xy) = /(at) + /(y) (x > 0, ? > 0) 

(iv) f(xy) = f(x)f(y) (x > 0, y > 0) 

for which the derivative of/ exists and is continuous wherever /is 
defined. For (ii) and (iv) you may assume that /never takes the value 
zero. 



15 



POWER SERIES 



15.1 Interval of convergence 

A power series about the point % is an expression of the form 

in which x is a variable. 

15.2 Theorem The set of values of x for which a power series 

converges is an interval with midpoint £. 

Proof We shall show that, if the power series converges when x =y, 
then it converges for all x satisfying \x—$\<\y—%\. The theorem then 
follows. 

I.v-£i Ir-fl 

■< 5«-« »- 



«- 



*r 



If the power series converges when x =y, then 
a n (y — %)" -*• as n -> °° (theorem 6.9). 
Hence the sequence <a n (j' — £)"> is bounded (theorem 4.25). Thus, for some H, 
\a n {y-%) n \<H in = 1,2,...). 
Now suppose that \x — 1|< I y — f I. Then 
I* -« 



P = 



Hence 



y-ll 



<1. 



\a n (x-t-)"\ = l^^-O^.P^-^P" (« = 1,2,...). 
The convergence of the power series 



143 



144 



Power series 



I fl„(jf-0" 



now follows from the comparison test (theorem 6.15) since 2„=oP" converges. 

We call the set of values ofx for which a power series converges its interval of 
convergence. If the endpoints of the interval of convergence are £ — R and 
| + R, we call R the radius of convergence of the power series. (If the interval of 
convergence is the set U of all real numbers, we say that the radius of conver- 
gence is infinite.) 

The proof of theorem 1 5.2 shows that a power series converges absolutely at 
all points of its interval of convergence with the possible exception of the end- 
points. At the endpoints the series may converge absolutely or it may converge 
conditionally or it may diverge. 

It is sometimes useful to observe that the radius of convergence R is given by 



- = limsup |a„| 1/n 



and 



R 



= lim 



The formulae are justified by propositions 6.17 and 6.18. In both cases, appro- 
priate conventions must be adopted if the right hand side of the formula happens 
to be or + °°. In the second case, of course, the sequence <|« n+1 \j\a„ |> may 
oscillate. If this happens, the limit does not exist and so the second formula 
becomes useless. 



15.3 Example We list some power series about the point 0, together with 
their intervals of convergence. In each case, the radius of convergence can be 
calculated using the second of the formulae above. In the last three cases one is 
then left with the problem of whether or not the endpoints of the interval of 
convergence belong to the interval of convergence. 

Power series Interval of convergence 



I 

n=o 



I inx) n {0} 

n=0 



Z* n 



(-1,1) 



Power series 



145 



:Ti n 



[-1,0 



[-1,1] 



15.4 Taylor series 

Suppose that /can be differentiated as often as we choose in some open 
interval / containing the point g. Then its Taylor series expansion about the 
point £ is 



f ^-j^/ (B) tt) = my 



fe=W*=* /»+... 



It should not be automatically assumed that this power series converges with 
sum/(x). Indeed there is no reason, in general, to suppose that it converges at 
all (except when x = %) and, even if it does converge, its sum need not be f(x) 
(see exercise 1 5 .6(6) below). A function / which is the sum of its Taylor series 
expansion in some open interval containing £ is said to be analytic at the point £. 

The most natural way of showing that a function is analytic is to prove that 
the error term in Taylor's theorem (theorem 11.10) tends to zero as n ■* <*>. 



75.5 Example It is paTfeularly easy to write down the Taylor series expan- 
sion of the exponential function about the point 0. If f(x) = exp x, then 
f in \x) = exp x. Since exp = i , the expansion is 



«• x 

y — 

From examples 1 5.3 we know that this power series converges for all x. Is its 
sum e x l From Taylor's theorem we know that 



x x 2 X*- 1 

»* = 1 + — + — + . . . H 

1! 2! (h- 1)! «! 



+ x - n 



where 77 lies between and x. Hence 



x x 

1 + — + — 4- . . .+ 
1! 2! 



*~] 


= 


x" 
n\ 




(« - l)!j 






n\ 




■» as n -* °° 



146 



Power series 



(by exercise 4.20(4) or by theorem 6.9). The partial sums of the power series 
therefore converge to e x , and thus, for all values of x. 



= 1 + — + — + — +. . . 
1! 2! 3! 



15.6 Exercise 

(1) Determine the intervals of convergence of the following power series 



» £ e 



00 I nix" 

n=0 



OH) I (-1)" 



(2«>! 



0v) I C-I) n : 



(«!) 



n 2 



(2« + 1)! 



(y) L ^rr.x n (vi) E 
& (2b)! 



n=l V" 



(2) Prove that the Taylor series expansion of log (1 + x) about the point 
is 



I (- ir 1 



n=l 



X* X X 

= x + — + ... 

2 3 4 



Using Taylor's theorem in the form given by exercise 13.26(6) show 
that this power series converges to log (1 + x) for — 1 < x < 1 . Hence 
give another proof of the identity 

log2 = l-i + i-|+... 

(see exercise 14.3(5)). 
(3) Prove that the application of Taylor's theorem in the form of theorem 
11.10 (with | = 0) to the function log (1 + x) yields a remainder term 
of the form 



E« = 



(-1)' 



1+7, 



where r? lies between and x. For what values of x is it possible to 
show that E n -*■ as n -* <x> without further information about the 
manner in which 77 depends on x and ;i? 
(4) Prove that 



1 



1 



(« + 1)! (« + 2)! 
Deduce that 



...< 



1 



K + 2 



(n + 1)! ' n + 1 ' 



Power series 



147 



and hence show that 2-7083 < e < 2-7184. 
*(5) Prove that e is irrational. [Hint: assume that e = mjn, where m and n 

are natural numbers and seek a contradiction using the inequality of 

question 4.] 
*(6) Let/: |R-*R be defined by 



fix) = 



,-l/x" 



(**0) 

[0 (x = 0). 

Show that, for any x J 2 0, 



-l/x' 



/<">(*) = P n H e 

where Pis, a polynomial of degree 3m. [Hint: use induction.] Using 
exercise 14.5(2), deduce that, for each natural number n, 

^-*0asx-»-0+. 

x 

Hence show that /can be differentiated as many times as we choose at 
the point and/ (n) (0) = (n - 0, 1 , 2, . . .). 

Write down the Taylor series expansion of/ about the point 0. For 
what values of jc does this converge to/(x)? 



15.7 Continuity and differentiation 

The following proposition is useful. Its proof has been placed in the 
appendix since it is somewhat involved. The reader will encounter the proof 
again at a later stage when studying 'uniform convergence'. 

15.8 Proposition Suppose that the power series 

f( X ) - 1 4&-£r 

n=0 

has interval of convergence /. Then its sum is continuous on / and differentiable 
on / (except at the endpoints). Moreover 



fix) = £ m£x-Qr*. 



148 Power series 



15.9 Example Proposition 15.8 allows us to tackle exercise 15.6(2) in an 
alternative way. 

We know that the power series 

I(-l)"-'- 
n = i " 

has interval of convergence (— 1 , 1 ] (see Example 1 5.3). Hence, by proposition 
1 5.8, it is differentiable on (— 1 , 1) and 



d K-D"- 1 



n-l„n-l 



= £ (- I)"" 1 * 



= 1 -x+x 2 — x 3 + .. . 



1 



1 +x 
It follows that, for each x 6 (— 1 , I ), 



(-!<*<!). 



D log(l+*)- E (-!)"- 



l+x l+x 



= 



and therefore that 



log(l+.x)= £ (-I)"" 1 — +c (-KK1) 

where c is a constant (theorem 1 1.7). By substituting x = 0, we see that the con- 
stant c — 0. 

Since the power series is continuous on (— 1, 1], 



log 2 = lim 



1ob(1+jc) 



= lim I (-!)»-»- 

x ->i- \ n=l n 



= I 



(-ir 



15.10 Exercise 

(1) Let a be a real number. By considering 

D L +x) -« l «(*-». ..<a-n + l) x „ 



{ 1=0 

for \x | < l.show that 



Power series 



149 



(1 + x) a = 1 + ax 



a ( a ~ 1 ) ..- 
2! 



x 2 + . . . 



provided that | X | < 1 (general binomial theorem). 
(2) Prove that the power series 



Z n-x 



2„n 



n=0 



converges for | x \ < 1 and determine its sum. 

(3) Suppose that the power series 

f( X ) = | a n x n 

n=0 

has interval of convergence /. If y E I but is not an endpoint of/, 
prove that 

\jMdx= 1 -Shy*. 

Jo n=0 n 4- 1 

(4) Prove that 






i log ( 1 -x) 



dx. 



n=0 n 

(5) Suppose that the two power series 

n=0 «=0 

converge in some open interval /containing \. Prove that their sums are 
equal for each x G / if and only if 

a n m b n (n m 0, 1,2, . . .). 

(6) Suppose that the power series 

y = fix) = t On** 

n=o s — - 

converges for all real x and satisfies the differential equation 
dy 



dx 



= y. 



l 



Show that a n+l = — — a n (n = 0, 1 , 2, . . .) and deduce that 
n + 1 

f{x) - a e*. 



16 



TRIGONOMETRIC FUNCTIONS 



16.1 Introduction 

We based our definitions of the exponential and logarithm functions on 
the formula 

f * dt 
We could equally well have begun with the differential equation 

%-y 

dx 

and defined the exponential function as the sum of a power series as indicated in 
exercise 15.10(6). 

In a similar way, the definitions of the trigonometric functions can be based 
on the formula 



arctan x 



-r 

Jo 



dt 



]+t 2 ' 
We prefer, however, to base them on the differential equation 



\ + y = 0. 
dx* 

We already have some intuitive ideas about the sine and cosine functions from 
elementary geometry and trigonometry. It is therefore sensible to begin by indi- 
cating why the sine and cosine functions, as conceived of in trigonometry, 
should be expected to satisfy this differential equation. 

In the diagram, the angle x is measured in radians and.y = sin x and z = cos x. 

By the sort of argument considered adequate in elementary trigonometry, we 

obtain 

dy , dz 

v = sin x = 

dx 



= —z. 



150 



z = cosx = — : ; 
dx 


y = sinx = 


ich it follows that 




d 2 y _ dz 

dx 2 ~ dx ~ y '' 


d 2 z dy 

dx 2 " dx 






Trigonometric functions 



151 




16.2 Sine and cosine 

Suppose that the power series 

/(*) = t **" 

n=0 

converges for all real values of x and that 

/"(*)+/(*) = o. 

By proposition 15.8, 

f(x) = a + a t x+ a 2 x 2 +...+ a n x n +... 

fix) = «!+ 2a 2 x+ 3a 3 x 2 +...+ in + l)a nti x n + . . . 

fix) = 2a 2 + 3 . 2a 3 x + 4 . 3a A x 2 + . . . + (n + 2)(n + l)a n + 2 x" + ... 

Thus, to satisfy the differential equation, a Q and a, may be chosen in any 
way, but then the remaining coefficients must be chosen to satisfy 

in + 2)in + l)a n+2 + a n = (« = 0,1,2,...). 

This recurrence relation is readily solved and we obtain 



„2n 



,,2n+l 



'*I<- I >*e* +8 'S ei) V+»>' 



n=0 



We want sin = and D sin = cos = 1 . We therefore define 
sin* ■■ £ (-1)" 
Similarly 



n=0 



x 2n+l =;t _^! + 

(2n + l)! * 3! 



^2" 



««-£<-»" m 



1 - 



?! 



152 Trigonometric functions 



Both these power series converge for all values of x by comparison with the 
power series for e x . 

Note that we take these formulae as definitions. The definitions of elementary 
trigonometry concerning the 'opposite', the 'adjacent' and the 'hypotenuse' are 
not precise enough for our purposes in this book. The remarks of §16.1 are 
simply intended to indicate why we chose to define sine and cosine as we did. 



16.3 Exercise 

(1) Prove the following. 

(i) cosO = 1 (ii) sinO = 

(iii) cos (— x) = cos x (iv) sin (— x) = — sin x. 
Show that, for all values of X, 
(v) D cos x = — sin x (vi) D sin x = cos x. 

(2) Let y be any real number and define 

g(x) = sin (x + y) — sin x cos y — cos x sin y 

h(x) = cos(x +y)~ cosx cosj' + sin* sin.y. 

Differentiate {g(x)} 2 + {h(x)} 2 with respect to x and hence show that, 
for all values of* andy, 

sin (x+y) = sin x cos y + cos x sin y 

cos (x + y) = cos* cosy — sin x sin y. 

(3) Use the previous question to prove the following. 

(i) cos 2 x + sin 2 * = 1 (ii) | cos x | < 1 and I sin x I < 1 

(iii) sin 2x = 2 sin x cos x (iv) cos 2x = cos 2 x— sin 2 x. 

(4) Use L'Hopital's rule to prove that 

„, sin x 

(i) »1 asx^-0 



... 1 — cos x 1 
00 — ^ >-asx 



0. 



trt 



(5) Use the mean value theorem to show that, for all values of x 

sinx 

<1. 



Hence show that the series 2™ =1 sin (l/« 2 ) converges. Discuss the con- 
vergence of S" =I sin (l/«). 



Trigonometric functions 



153 



*(6) Use exercise 13.26(5) to prove that 
c" sin t 



-dt 



m 



provided that n>m> 0. Explain why it is possible to deduce the 
existence of the limit 
"sin t 



(-"sin 
lim — 

„->=oJl t 



dt. 



16.4 Periodicity 

Since cos = 1 and cos (— x) = cos x and the cosine function is con- 
tinuous, there exists a % > such that cos x > for x S (— £, £). 

But the cosine function is not positive all the time. If this were the case, then 
it would follow from the formula D 2 cos x = — cos x that the cosine function 
was concave. But a bounded, differentiable function cannot be concave unless it 
is constant (exercise 12.21(5)). 

It follows that there exists a smallest positive number % for which cos % = 0. 
We define the real number n by 
\n = %. 

By definition, cos £tt = cos (— \n) = and cos x > for - \v <x < \-n. We 
show that sin \it = 1 . It follows from the formula cos 2 x + sin 2 x = 1 that 
s i n 2 Jtt = 1 . But D sin x = cos x and so the sine function increases on 
[— {-n, j7rj . Since sin = it follows that sin jtt > 0. 

We now appeal to the formulae of exercise 16.3(2). We have 

sin (x + j7r) = sin x cos \ti + cos x sin jt7r = cos x 

cos (x + jit) = cos x cos \tt — sin x sin \tt = — sin x 

from which it follows, in turn, that 

sin (x 4- 2n) = sin x 

cos (x + 2n) = cosx. 

Because of these last formulae we say that the sine and cosine functions are 
periodic with period 2n, 




154 Trigonometric functions 



16.5 Exercise 

(1) Prove thai the cosine function decreases on [0, rr] and increases on 
[ n, 2n] . Prove that it is concave on [— \-n, \tt} and convex 

(2) We define the tangent function by 



on [2^. jt] 



tan* = 



sm x 

COS X 



(x^{n + \)rt:n = 0,±l,+2, . . .). 



Prove that tan (x + 7r) = tan x provided that x =£ (« + £) i\ where n = 0, 
±1, ±2, . . . Show that the tangent function is strictly increasing on 
(— \ir, \tt) and that tan x •* + °° as x -> \tt — and tan x -* — °° as 

(3) Show that the sine function is strictly increasing and continuous on 
[ — ?"■> 5 7r ] ar| d tnat { ' ie image of this interval under the sine function 
is [— 1, 1]. If we ignore the fact that the sine function is defined out- 
side the interval [— \-n, |ir], the function we obtain therefore admits 
an inverse function with domain [—1,1] and range [— \ti, \-n\ . We 
call this function the arcsine function. Draw a graph of this function 
and calculate D arcsin x for — 1 <x < 1 . (Some authors use the 
notation sin" x, but we prefer not to, since the sine function has no 
inverse. The arcsine function is the inverse of the restriction of the sine 
function to [— \-n, ^n]). 

Discuss the arccosine function obtained as the inverse of the restric- 
tion of the cosine function to [0, rr\ . Calculate D arccos x and explain 
how the arcsine and arccosine functions are related. 

(4) Show that the arctangent function, obtained as the inverse of the' 
restriction of the tangent function to (— \tr, \-n), has domain U and 
range (— j7T, \tj). Draw a graph and calculate D arctan x. Hence show 
that 



dx 



Jo \+x 2 2' 
+ (5) Suppose that /has a continuous derivative on R and that 



/M+/O0 =/ 



x + y 

\—xy 



for each x andy such that xy <\. Prove that, for some constant C, 
f(x) - C arctan x. By writing* =jc in the formula and considering 
what happens when x -*• 1 — , prove that arctan 1 = \n. 
Show that, for - 1 <x<l. 



X X X 

arctan x = x 1 K.. 

3 5 7 



Trigonometric functions 



155 



and hence obtain the identity 



J„ = !_«+£->+... 



t 



(6) Let/: 



fix) = 



*U be defined by 

sin - {x # 0) 
x 

(x = 0). 



Let#: R -> R and/i: R->R be defined by g{x) = xf(x) and 
h(x) = x 2 f(x). Draw graphs and prove the following. 

(i)/is not continuous at the point 0. 

Qi)glS continuous at the point but not differentiable there, 
(iii) h is differentiable at and h'(Q) = 0. 






17 



THE GAMMA FUNCTION 



17.1 Stirling's formula 

If ( a n ) and < b„ > are sequences of real numbers, the notation a n ~ b n 



means that 



*■ 1 as n -+ ca . 

b„ 



17.2 Proposition {Stirling's formula) 

nl~y/(2ir)n n n i/2 e- n . 

Proof Consider the sequence < d n > defined by 

d n = log «!—(« + j) log n+n. 
We seek to show that (,d n > decreases and so we examine the sign of 

d n -d n+l = -log(«+ 1) - (« + |) log « + (n + |) log (« + 1) — 1 

4 4-4^-' 



2n + l |l+(2n+l) _1 



But, for |jc|<1, 



1 



m = ^io g 



i +* 



1 -x 

2 



-1 



2*11 2 3 



jc 2 x A x 6 

3 5 7 



2 3 



-1 



0) 



In particular, f(x)>0 for U|< 1 and thus d„ ~d n+1 >0 (n = 1, 2, . . .). 
Therefore <c/„> decreases. 



756 



The gamma function 
From (1) it also follows that, for | x \ < 1, 
f(x)<j{\+x 2 +x*+...} 



157 



Thus 



3(1 -x 2 ) 



2\" 



rf n -"<*„+,<; 



1 



3{(2n + l) 2 -l} 12« 12(n+l) 
It follows that the sequence (,d n —(12b)" 1 } is increasing. In particular, 
d n -{\2n)- i >d l --h (n = 1,2,...) 

and hence the sequence < d n > is bounded below. Since < d n > decreases, it follows 
that {d n ) converges. 

Suppose that d n -* d as n -*■ °°. Since the exponential function is continuous at 
every point, exp d n •* exp d as n •*. •*>, Let C = exp d then 



n\ 



n"n U2 e- n 



C as n -*■ °°. 



It remains to show that C = V(2t)- We leave this as an exercise (exercise 
17.4(1) and (2)). 



1 7 .3 * The gamma function 

The gamma function is defined on (0, °°) by the formula 



r(x) = J~*V -l <r*«a («>o). 



To justify the existence of the improper integral, we appeal to proposition 13.29. 
In view of the inequality 

the existence of the integral £ fl t x ''e-'dt follows from that of £ f x_1 rfr, pro- 
vided that x > 0. Since r x+1 e"' ■* as t -> + °°, we have, for some i/ >0, 

f*-*«-»<JSft** (r>l) 
and hence the existence of r*°° t x ~ l e~'dt follows from that of/,"*" f " 2 dr, 



/7.4 + Exercise 

t(l) Let / n = /f" 3 sin" xdx (n = 0, 1 , 2, . . .). Prove that < /„ > is a decreasing 
sequence of positive numbers which satisfies nl„ = (n — l)/ n -2 
(n = 2, 3, . . .). 



158 Tlie gamma function 



Deduce thai 



hn 



1 as n -> °°. 



^(2) With the notation of the previous question, establish the identities 
(2n)\ n 



hn (2 n n\f 2 
(2"«!) 2 



'2n*l 



(2n+ I)! 



(" =0,1,...) 
(« = 0,1,...). 



Hence show that the constant C obtained in §1 7.2 satisfies C = \/(2n). 
T (3) Let d n = log n\ - (n + j) log n+n. Show that the sequence 

{d n — (12«) -l > increases but that the sequence (d n — (12n + 1) _1 > 
decreases. Deduce that 



g l/(12n-H)^ 



n\ 



\/(27r)n"n V2 e- n 



<P vl2n 



Hence estimate the error on approximating to 100! by Stirling's 
formula. Is this error large or small compared with the value of 100!? 
+ (4) Show that r(x + l) = xr(x)(x>0). Deduce that, fom = 1,2,3, ... , 

r(n + l) = b! 

' (5) Prove that the gamma function is continuous on (0, °°). [Hint: If 

0<a<a<.K<_)><£<|3, prove that, for some constant H which does 
not depend on x aty, \r(x) - T(y) \<ff\x —y | {r(a) + 1X0)}.] 

■ (6) Prove that the logarithm of the gamma function is convex on (0, °°). 
(Hint: exercise 12.21(6) and theorem 13.25.] 



1 7.5" 1 Properties of the gamma function 

The gamma function provides a generalisation of the factorial function 
(see exercise 17.4(4)). Is it the only such generalisation? 

1 7.6* Theorem Let /be positive and continuous on (0, °°) and let its 
logarithm be convex on (0, °°). If/ satisfies the functional equation 

/(x+1) = */(*) (x>0) 

and/(l)= l.then 

/(*) = T(x) (x>0). 

Proof The proof consists of showing that, under the hypotheses of the 
theorem, for each x > 0, 



The gamma function 



fix) = lim 



n x n\ 



„"'- x(x +])... (x+nY 



159 
(2) 



It follows from exercise 17.4(4, 5 and 6) that the gamma function satisfies 
the hypotheses of the theorem. Since a sequence can have at most one limit, we 
can therefore conclude from (2) that/(x) = T(x) (x > 0). 

The proof of (2) uses the convexity of log / Suppose that s < f < s + 1 . Then 
we may write t = as + (3 (x + 1) where a>O,0>Oanda + |3=l. Now 
t = (a + 0)s + (3 = s + and so = t - s. 

From the convexity of log/, it follows that 

log/(r) < a log/(x) + log/(s + 1) 

fit) < imrm + o}" 

= s^m = s<-*m. (3) 

Since s < t < s + 1 , we also have / — I < s < f . Making appropriate substitu- 
tions in (3), we obtain 

f(s) < (t - ir**V(* ~ 1) - fr~ l) s "'/(0- w 

Combining (3) and (4) yields the inequality 

(t-\y- s f( S )<f(t)<s'- s f(s). 

Now suppose that <x < 1 and that n is a natural number. We may take 
s = n + 1 and t = x + n + 1 . Then 

(x + n) x f(n + 1)< f(x +n + 1)<(» + D x /(« + 1 )• (5) 

From this inequality it follows that 

(x + n) x n ! < (x + n)(x + n - 1) . . . xf{x) <(n + l) x n\ 

,^ >+.x»+ ; .-n...v M < (, + i)', 

This completes the proof of the formula (2) in the case when < x < 1 . The 
general case is easily deduced with the help of the functional equation 
f(x+l)=xf(x). 



17.7^ Exercise 

+ (1) lfx>0, prove that 



r« = r 



log- 

I t 



dt. 



1 60 The gamma function 

+ (2) Prove that T(x) ~ yj(2n)x x x' in e' x . This is Stirling's formula for the 

gamma function. [Hint: see theorem 17.6, inequality (5).] 
*(3) Use L'Hopital's rule to show that 



lim 

z-»0 



log(l +z)-z 



With the help of Stirling's formula for the gamma function, deduce 
that, for all x > 0, 



(V«)/u(« +*>/«)• 



1 



V(2rr) 



e ' as u 



where 



1 



AM - j^-V. 

* (4) The beta function is the function of two variables defined for x > and 
y > by the formula 



B(x,y) =P' ^-'(l-ty-'dt. 
J-*o 



Check that the improper integral exists provided that x > and 
y > 0. Prove that, for a given fixed value of y > 0, B(x, y) is a positive, 
continuous function of x on (0, °°) whose logarithm is convex on 
(0,-). 
'(5) With the notation of the previous question, prove that, for a given fixed 
value of y > 0, the function /: (0, °°) -*R defined by 

f(& = r( * ( + y f B(x,y) (x>0) 

satisfies the conditions of theorem 17.6. Deduce that, for x > and 
^>0, 



B(x,y) = 



r«ro>) 



r(x+y) 
T (6) Use the previous question to evaluate T(j). Hence show that 

P~ e~ x2 ' 2 dx = V2ir. 



18 



APPENDIX 



18.1 Introduction 

In the preceding chapters a number of results were stated without 
proof. These results were referred to as 'propositions'. The proofs were omitted 
from the main body of the text to avoid confusing the issue with too much 
detail. Instead we give the proofs here. 



1 8.2 Bounded sets 

Proposition 2.3 A set S of real numbers is bounded if and only if there 
exists a real number K such that | x \ < K for any xES. 

Proof By exercise 1.20(1), \x | <^if and only if -K<x<K. If 
| x | < K for any x e S, it follows that - K is a lower bound for 5 and K is an 
upper bound for S. Thus 5 is bounded. 

Suppose, on the other hand, that S is bounded. Let H be an upper bound and 
h a lower bound. Put K = max {| H I, I h !}. Then 

-K<h<x<H<K (xGS) 
and hence | x \ < K for any x G 5. 

1 8 .3 Combination theorem 

Proposition 4.8 Let x n -» / as n ■* °° and y n ■* m as n -* °°. Let X and m 
be any real numbers. Then 

(i) Xx n + p.y n -*\l + /t»i asn-*<* 
('0 Vn -*lm as n-*°° 

x I 

(iii) — -> — as n -* °° (provided that m i= 0). 

y n m 

Proofs o/0'O and (iii) 

(ii) Since <x„) converges, it is bounded (theorem 4.25). Suppose that 
!*„!<*(» = 1,2, 3,...). Consider 



161 



162 Appendix 



\x n y n -lm\ = \x n y n -x n m+x n m-lm\ 

< I x n I • I y n ~~ m I + I m I . I x n — 1 1 (triangle inequality) 

< K.\y n -m\ + \m\.\x n -l\ 
= z n - 

But x„ -► / as « -►o and so | x„ — l\-*Q as n -►<*> (exercise 4.29(H)). Similarly 
| y n — m | -* as n -* °°. From proposition 4.8(f), z„ -*0 as/i -»•<». Hie result 
then follows from corollary 4.1 1. 

(iii) Since y„-*-m as« -*■**, \y„ | -> \m | as« -*°° (exercise 4.29 (lii)). Since 
m # 0, | m | > 0. It follows from exercise 4.29(2) that we can find an N such 
that, for any n >N, 

\y n \>h\m\. 

For n > N, consider 



X n_J_ 

y n m 



'nx n -yj 



my„ 



< 



m 



\mx n -y n l\. 



By proposition 4.8 (i), mx n - y n l -*■ ml — ml = as n -* °°. The result therefore 
follows from corollary 4.1 1 . 



18.4 Subsequences 

Proposition 5.13 Let (x n > be a bounded sequence and let L be the set 
of all real numbers which are the limit of some subsequence of (x n ). Then L has 
a maximum and a minimum. 

Proof The set L is non-empty (Bolzano-Weierstrass theorem) and 
bounded (theorem 4.23). Hence L has a smallest upper bound/. We seek to show 
that IGL, i.e. there exists a subsequence (x n > such that x n -*lasr-*°°. 

Let e > 0. Then je > 0. Since 7 is the smallest upper bound of L, 1 — £e is not 
an upper bound of L. Hence there exists znlEL such that 

-I, 



l>l>l-\e. 

and therefore 

|f-7|<ie. 

Because / G L, we can find a subsequence < x m > which satisfies x m -*■ I as r - 
Hence there exists an R such that, for any r >R, 

\x m — u<y. 



(i) 



(2) 



Appendix 163 

Taking (1) and (2) together, we obtain, for any r >R, 

\x mr -i\ = \x mr -i+i-h 

< l* mr -/| + |/-7l 

< ke + he 
i.e. for any r >R, 

\x mr -l\<e. 

This does not conclude the proof since the subsequence <x mj _) will change as e 
changes. All we have shown is that, given any e > 0, there exists an infinite 
collection of terms x n of the sequence < x n > which satisfy 

|x„-?|<e. ( 3 > 

With this information, we construct inductively a subsequence (x„ r ) which 
satisfies x n -* 1 as r •* °°. 

Take e = 1 in (3). Then there exists an «, such that | *„_ - / 1< 1 . Take 

e = \ in (3). Then there exists an n 2 > n , such that | x„ 2 - 1 1 < k- In this wa Y we 
construct a subsequence <jc„ > which satisfies 



l*n r -'l<7 

and hence x„ r -*■ 7 as r -*■ °° (sandwich theorem). 

A similar argument shows that the infimum / of L also belongs to L. 

Proposition 5.17 Any convergent sequence is a Cauchy sequence. 
Proof Let e > be given. Then £e > 0. If x n ■* I as n -* °°, then we can 
find an A' such that, for any n>N, 

\*n-l\<te. 
Equally, for any m > A^, 

\x m -l\<i L e. 
Thus, if m > N and n > N, 

\x n -x m \ = \x n -l + l-x m \ 

< \x n -l\ + \x m -l\<ke + he = e 
and thus ix n ) is a Cauchy sequence. 

Proposition 5.18 Any Cauchy sequence is bounded. 

Proof Let <x„> be a Cauchy sequence. It is true that, for any e > 0, we 
can find an N such that, for any n > N and any m > N, | x n - x m | < e. In parti- 
cular, this is true when e = 1 , i.e. there exists an N t such that, for any n>Ni 



164 Appendix 

and any m>Ni. 

\x n -x m \<\. 
Take m = N t + 1 . Then, by theorem 1 . 1 8, for any n>N lt 

\x n \-\x Ni + l \<\x n -x Nitl |< 1 

l*nl<l*JVll + !■ 

Now take 

A' = max {|*i | 7 |x 2 1, . . . , \x Ni |, |*n,-mI + 1} 
and it follows that | *„ | < A" (w = 1 , 2, . . .). 



1 8.5 Tests for convergence of series 

Proposition 6.17 {ratio test) 

Let S~ = , a n be a series which satisfies 



lim 



<n+l 



= /. 



If /> 1, the series diverges and, if / < 1, the series converges. 

Proof The text indicates the proof for the case / < 1 . If / > 1 , we may 
take e > so small that / — e > 1 . Then, for a sufficiently large value of /V, 



Kv-h 



«n-l 




"n-2 




"N+2 



>Q-e) 



n-N*l 



a N+i I -*• + °° as M 



Hence a n j*Q as n -*■ °° and so 2~=i a n diverges. 

Proposition 6.18 (nth root test) Let 2„ =! a„ be a series which satisfies 
lim sup |a n | 1/n = /. 

If / > 1 , the series diverges and, if / < 1 , the series converges. 

Proof Let x n = \a n |"". If / < 1, we may choose e > so small that 
' + e < 1 . From exercise 5.15(4) it follows that we can find an iV such that, for 
any n > N, 



i.e. 



x n < I + e 
\a n \ <(/ + £)". 



Since 2™ =1 (/ + e)" converges, the convergence of E~ =1 a n follows from the 
comparison test. 



Appendix 



165 



If I > 1 , we may choose e > so small that / - e > 1 . Let < x„ r > be a sub- 
sequence such that *„ r -> I as r -*■ °° (proposition 5.13). We can find an R such 
that, for any r>R, 

x„>Q-e) 

\a nr \>(l-e) n r-> + °°asr^°°. 
Hence a„^0asn-» M (theorem 5.2) and thus S*-i a n diverges. 



1 8.6 Limits of functions 

Proposition 8.4 Let /be defined on an interval (a, b) except possibly 
at a point g e (a, ft). Then/(*) ■* I as * ■+ g if and only if /(x) -Was* ■* | -and 
/(*)->/ as x->£+. 

iVoo/ (i) Suppose that /(*) -» / as * -*• |. Let e > be given. Then we 
can find a 5 > such that 

|/(*)-/|<e 

provided that 0<|*-£l<5.But£-5<*<£ implies that < I * - % |< 5 
(see §8.3). Hence | - 6 < * < g implies that | /(*) - / 1< e. Thus /(*) ■* I as 

Similarly, % < * < g + 5 implies that < I * - g 1< 8. Hence ?<*<? + 5 
implies that | /(*) -l\<e. Thus /(*) -* / as * -* g +. 

(ii) Suppose that /(*) -* I as * -*■ % - and /(*) ■* I as * ■+ g +. Let e > be 
given. We can find a 5j > such that | /(*) - / ] < e provided that 

g-5 1 <*<|. 
Also we can find a 6 2 > such that | /(*) - / 1 < e provided that 

£<*<g + 8 2 . 
Let 5 = min {S, , 5 2 } . Then | /(*) -l\<e provided that 

£-6<*<£andg<*<g-6 



i.e. 



i.e. 



^-5<*<? + 5and*^? 



0<|*-£l<5. 
Thus /(*)-* /as*-* |. 

Proposition 8.12 Let /and g be defined on an interval except possibly 
at £ G (a, ft). Suppose that f(x)-+l as * ■* g and g(*) ■* m as * -» g and suppose 
that X and n are any real numbers. Then 

(i) X/(*) + pg(x) -> X/ + (WW as * ■* £ 

(ii) /(*)£(*) ^/« as* ^£ 



166 Appendix 



(iii) f(x)/g(x) -* l/m as x -*■ £ (provided w =£ 0). 

Proof (i) Let <x n > be any sequence of points of (a, b) such that x„ =£ £ 
(n = 1 , 2, . . .) and x„ -> £ as n -> <». By theorem 8.9, /(x„) -* / as n -*■ °° and 
g(x n ) -* m as « ■+ °°. By proposition 4.8(i) 

*/(*n) + Atf(*n) -* X/ + /im as « -> °°. 
Appealing to theorem 8.9 again we obtain 

X/(x) + Mf (■*) ~* M + A"" as* •* f. 
Items (ii) and (iii) are proved in exactly the same way. 

Proposition 8.14 Let /, g and h be defined on (a, b) except possibly at 
£ G (a, b). Suppose that g(x) -*■ I as x -*• £, h (x) -> / as x •+ £ and that 

g(x)<f(x)<h(x) 

except possibly when X —%. Then fix) -*■ I as x -* §. 

Proof Let < x„ > be a sequence of points of (a, £) such that x„ =£ £ 
(n = 1,2, . . .) and x„ ->• £ as n ->• °°. By theorem 8.9,#(x„)->/ as «-*» and 
/i(*n) -> Z as n -* oo. Since g-(x„) </(x„) < /z(x„) it follows from theorem 4.10 
that/(#„) -> / as n -*■ °°. Hence /(x) ■* / as x ■* £ by theorem 8.9. 



18.7 Continuity 

Proposition 9.3 Let /be defined on an interval. Then /is continuous 
on / if and only if, given any x G / and any e > 0, we can find a" 5 > such that 

l/(*)-/O0Ke 

provided that y& /and \x— y | <5. 

Proof If *€/ but is not an endpoint of/ then the condition y G / and 
I x —y | < 5 is just the same as | x — y |< 5 provided that 8 is sufficiently small. 
The criterion given in the proposition therefore reduces to the assertion that 
f(y) ""*■/(*) as y •* x, i.e. /is continuous at x. 

Suppose that x € / and is a left hand endpoint of /. The the condition y G I 
and | x — y | < 5 reduces tox<y<x + 8, provided that 5 is sufficiently small. 
The criterion of the proposition therefore reduces to/(^) ~*f(x) asy -*x +, 
i.e. /is continuous on the right at x. Similarly if x G/and is a right hand end- 
point. 

Propositions 9.4, 9.5 and 9.6 are entirely trivial consequences of the results 
indicated in the text. 



Appendix 

5 1 8.8 Integration 

Proposition 13.4 is proved in §13.2. 



767 



Then 



have 



Proposition 13. 7 Let /be continuous on [a, b] and let c be constant. 

\\ {/(<) + c} dt = £ f{t) dt +c(h- a). 

Proof hztP= {yo,y\,y2, ■ ■ • >y«} denote a partition of [a,b]. We 



m\ 



U+c) _ 



inf {f(x) + c}=c+ inf {/(*)} = e + m\ n . 



y <x<y, j> <x<y, 

Similarly mj/ +c) = c + m^ (k - 1, 2, . . . >a). It follows that 

s if+c) (p) m | mg* e \y k -y k _ t ) 

fe=i 



n n 

= E wi'^fc-^fc-J + e Z (n-yfe-i) 

fe=l fe=l 

= S in (P) + c(b-a). 



It follows that 



\ {f(t) + c} dt = sup {S if * c \P)} = sup {S«\P) + cib -a)} 

Ja p p 

= sup {S<- n iP)}+ cib -a) = l" f(t)dt + c(b-a). 
p Ja 

Proposition 13.29 Suppose that is a function which is continuous 
and non-negative on the open interval /. Suppose also that /is continuous on / 
and satisfies 

|/(*)|<0(X) (xG/). 

If the improper integral of (p over the interval /exists, then so does that of/. 
Proof We consider only the case when / = (0, °°) and the integral 

/ = P°° <t>ix)dx 

exists. Let 

«n = f" /(*)<**; bn=r <Kx)dx (« = 1,2,...). 

Jn-l Jn-1 

Then the series 2~=i b„ is a convergent series of non-negative terms (with sum /). 
Also 



168 Appendix 

\a n \<b n (H = 1,2,...) 
and hence 2" =1 a n converges by the comparison test, i.e. 

■ N 



m = lim f f(x)dx 



exists. Given any X > 0, we may take N to be the smallest natural number satis- 
fying N>X. Then 



<&;v->-0as A 7 -*- 00 . 



It follows that 
-x 



= »? 



lim f f(x)dx = 
as required. 

1 8.9 Continuity and differentiation of power series 

Proposition 15.8 Suppose that the power series 
fix) = £ «„(*-*)" 

n=0 

has interval of convergence /. Then its sum is continuous on / and differentiable 
on / (except at the endpoints). Moreover 

f(x)= X mdx-HT' 1 - 

n=l 

Proof Let the radius of convergence of the power series be R. 

Suppose that X is any point of / other than an endpoint. We can then 
find another point x of / so that x lies between £ and x and thus 

[X-f|<|ar -||<fi. 



-/?— ■ x- 



-> 



-f— 1- 



£ x v„ 



From the formula of §15.1 



— = lim sup |a n l" n = lim sup 

K n~>°° n—*-°° 



n(n-\) 



l/n 




Appendix 

we may deduce the convergence of the series 
«(«-!) 



169 



I 



: a„*o" 2 



where Xo = *o — £• 

Put 5 = | x — x \. For values of y satisfying < | x — y | <6, consider 

^— x ^i n =i [ r— X 

where X = x — ? and F = y — if. We seek to show that A -* as Y-*-X. From 
exercise 3.11(2), 



n v"" 






- nX n ~ l = (Y"- 1 +Y"~ 2 X+ ... + YX"' 2 + X"' 1 ) - nl" 



= (Y"' 1 -X n ~ l ) + x(r n_2 - x"- 2 ) + ...+ X"-\Y- X). 

But from each of these terms we can extract a factor of (Y — X). The right 
hand side then reduces to the product of {Y — X) with the sum of \n(n — 1) 
terms of the form X r Y s where r + s = n — 2 . The number \n (n — 1 ) arises from 
the use of the formula 

1 + 2 + 3 + .. . + (n - 1) = %n(n - 1) (example 3.9). 
Since 

\X\ = |x-5l<|x -5l - I *,, I and m - ly-|KI*-tl + » » 1-Xbl 
we have 



Hence 



y n -z" 



r-AT 



-nJC 



n-I 



<-«(n-i}ijr i nH »|y-jfK 



lAKir-jriJ^i^Ti 



-+0as K-i-AT. 
It follows that /is differentiable at each x in / which is not an endpoint and 

/'(.*) = £ M#**. 
n-1 

There remains the question of left or right hand continuity at the endpoints 
of the interval of convergence /, if these happen to belong to /. A somewhat 
more subtle argument is required to deal with this question which is the subject 
matter of Abel's theorem quoted below. 

Abel's theorem If the series Sfe =0 a k converges, then 



1 70 Appendix 



lim £ a k x k \ = £ a k . 
*-+l- { fe=o ) k -° 

Proof Let e > be given. Since the series 2 a k converges, its sequence 
of partial sums is a Cauchy sequence. We can therefore find an N such that, 
whenever k>m>N, 



k 

Z a, 



<r- 



(4) 



Abel's lemma (which is easily proved by induction) asserts that 

+ v n . J u h . 



k-m k=m \\l=m 



We apply Abel's lemma with u k = a k and v k = x . Then 



k-m 



fe=m k=m 






It follows that, for n > m > N and < x < 1 , 



la k x k 

k=m 



<0-*)"f \ex k + \ex n 
k=m -> a 



3 (1 — x) 3 3 



We may conclude that 



I a k x' 

k=N 



< r- 



(5) 



Next observe that, for < x < 1 , 



£ a k x k - V a k 

fe=0 k=0 



N-i ! j 

< I \a n \(\-x") + -e + -e. 



fe=0 



This follows from (4) and (5). But the finite sum on the right hand side tends to 
zero as x -* 1 — . We can therefore find a 5 > such that, for any x satisfying 
\-8<x<\, 

N t\a n \(l-x")<\e. 
k=o 3 

Given any e > 0, we have therefore found a 6 > such that, for any x satis- 
fying 1 - 5 < x < 1 , 



T 



Appendix 



171 



_ a k x k - £ tffe 
fe=0 ft=o 



This concludes the proof. 



<e. 



18.10 Stirling's formula 

The proof of proposition 17.2 is completed in exercises 17.4(1 and 2). 



SOLUTIONS TO EXERCISES 



• Exercise 1.8 

1 There are three possibilities: x = 0, x > 0, x < 0. In the first case 
x 2 = 0. In the second case rule III yields that a; 2 = x.x >0.x = 0. 
Rewrite the third possibility iii the form >x and apply rule IV. Then 
= 0.x<x.x = x 2 . In each casex 2 >0, i.e.x 2 >0 or x 2 = 0. 

(i) We are given that < a < 1 . Since a > we may apply rule III. Then 
= 0. a<a.a<l.a = a. 

(ii) We are given that b > 1 . Since b > we may apply rule III. Then 
b 2 = b.b>l.b = b. 

2 Since bB >0, 

a. /4 

ff.fi = -.bB<-.bB = Ab (rule III), 
o fi 

We deduce that 

a{b + B) ■ a* + flfi < aft + v46 = (a + 4)4 (rule II). 

Since (6 + B)' 1 > and 6" 1 > (example 1.5), it follows from rule III 
that 

a a + A 
b b + B' 

The second half of the inequality to be demonstrated is obtained simi- 
larly. 

3 Suppose that a > b and c> d. Then, by rule U,a+c>b + c and 
b + c>b+d. Hence a + c>b + d (rule I). 

If b > and d > 0, we first observe that c > d > 0. Then, by rule 
III, ac > be and 6c > W, Hence ac > M (rule I). 

4 Substitute the values a = 5,b = 4, c = 3,d = 1 in inequalities (i) and 
(ii). We obtain the false assertions 



7 72 



Solutions to exercises (1.8, 1.12) 



173 



2 = 5-3>4-l = 3 



5 4 



i.e. 5 > 12. 



Note that the inequalities are false in spite of the fact that b >0 and 
d > 0. On the other hand inequality (iii) is true when b > and d > 
(exercise 1 .8(3)). However, if we take a = c = — 1 and b = d = — 2 in 
inequality (ii), we obtain the false assertion 

1 = (-l)(-l)>(-2)(-2) = 4. 

5 We have, for any e>0,b<a + e and a — e < b. Since b < a + e for 
any e > 0, it follows that b < a (example 1 .7). Since a < b + e for any 
e > 0, it follows that a < b. Hence a = b. 

6 Take x = (a + b)/2. 

• Exercise 1.12 

1 If n is an even natural number it may be expressed in the form n = 2k. 
But then 



x" = x 2h - 



= (x R ) 2 >0 



for all values of* (exercise 1.8(1)). Hence, if y < 0, x" = y has no 
solutions. The equation x" = has only the solution x = since x ¥= 
implies that x" =t 0. Uy > 0, our assumption about the existence of 
nth roots assures of the existence of a unique x>0 such that x" = y. 
But, since n is even, x" = y if and only if (— x)" =y. Hence the equa- 
tion has exactly two solutions, one positive and one negative. 

Next suppose that n is odd. If y = there is no difficulty in showing 
that*" =y has exactly one solution. Ifiy >0, there is exactly one 
positive solution and this is the only solution because x < implies 
.*" < when n is odd. If y < 0, we use the fact that z" = — y has one 
and only one solution and hence the same is true of (— xf = —y, i.e. 
x"=y. 




/ 74 Solutions to exercises ( 1. 12 J 



2 (i) 8 2 ' 3 = {(2 3 ) 2 }" 3 = {4 3 }" 3 = 4 

1/3 I . \ 1/3 

(ii)27 -4/3 = 



1 



27' 



1 



P 3 ) 4 J 



_1_ 

81 



(iii) 32 6/s = {(2 5 ) 6 } 1/5 = 2 6 = 64. 



3 We take for granted the truth of (i), (ii) and (iii) in the case when r and 
s are integers. To prove the results when r and s are any rational num- 
bers, it is helpful to have available the following preliminary results in 
which m and n denote natural numbers: 

(a)O m ) ,/n = (y Vn ) m (b)0 1/m )"" = y umn (c)y Un z Vn = (yz) Vn 

To prove (a), observe that 

{(}> "")"*}" = (y" n ) mn = {(y 1/ ")"} m = y m 

and hence (y v ") m is the unique positive solution ofx" = y m , i.e. 
(y »/»)"• = (y™) 1 '". To prove (b), observe that 

{(y 1/m ) 1/ "} m " = {{(y Vm y /n } n } m = (y" m ) m = y 

and hence (y Um ) Vn is the unique positive solution of x mn = y. To 
prove (c), observe that 

{y"» z >•'"}" = (ytoftfi-f = yz 

and hence y Vn z vn is the unique positive solution of x" =yz. 

We may now prove (iii), (ii) and (i) in the general case when r = pjm 
and x = qjn (p and q integers). 

(iii) yz r = ry)" m (z p ) ,/m = ov) 1 "" = &zff m 

= (yzf' m = (yzf 
00 Iff = {{(y p ) 1/m ]°} 1/ " = {l(y p )"} Vm } l,n = lyP-y""" 

_ „pq/mn _ ,JS8 
— > — >" 

(J) y r+s = y(P" + f")/mn _ r„pn qmil/mn _ (ypnyimnf qm\Vmn 

= yP""y<"" = y^ 

4 Write ax 2 + bx + c = a(x - a)(x - fi).Whena< x <0,jc -a>0 and 
x-p<0. Hence (x - a)(x - (3) < 0. When .v < a, x - a < and 

x - (3 < 0. and therefore (a: - a)(x - 0) > 0. When *>/3,x-a>0 
and x — > and therefore (* — a)(x — (3) > 0. 
'Complete the square' as in § 1.10 to obtain 

ax 2 + bx + c = {(2ax + b) 2 - (b 2 - 4ac)}/4a. 

Since (2ax + b) 2 >0, 






Solutions to exercises (1.12, 1.20) 1 75 

ax 2 + bx + c>c -\b 2 a' x 

with equality when lax + b = 0, i.e. x = —bjla. 

5 Apply the Cauchy-Schwarz inequality using the numbers \Ja\,\Ja 2 , 
. . . ,\Ja n and 1/Va,, . - • , l/y/a„. We obtain 

n 2 = I y/ai . —r- + ■ ■ ■ + V<* n • -j— 

<( ai +a 2 + ...+a n ) -+J+... + - 
\a, a 2 

and the result follows. 

6 t kk + Kf = I 4 + 2 Z a h b k + I 6? 



fe = i 



k=i fe=i fc=i 

1/2 / „ \l/2 



n 



n 



n 



fc=i \fc=i / \fc=i / fe=i 

(Cauchy-Schwarz inequality) 

LI + U.«) 

Consider, for example, the case n = 3 and suppose that (xj,x 2 ,x 3 ), 
(y lt y 2 ,y 3 ) and (z h z 2 ,z 3 ) are the co-ordinates of the vertices of a tri- 
angle in three dimensional space. The length of the side joining the firs', 
two vertices is given by 



\l/2 



fe=i 



< 



£ (x k -z k +z k -y k ? 
fe=i 

3 
fe = l 



1/2 



1/2 [,' U/2 



• Exercise 1.20 

1 We first show that \a\ < b implies -b < a < b and then that 
— fc < a < b implies \a\ < 6. 

(i) Suppose that \a\ < b. From theorem 1 .1 5, a < \a\ and \a\ >—a. 
Hence a <6 and — a< b, i.e. a >— 6. It follows that -fc< a < 6. 

(ii) Suppose that -b<a<b. Then a < b and -a < 6. Since, for each 
a, |a| = a or \a\ = —a, it follows that \a\ < b. 

2 From theorem 1.18, 



1 76 Solutions to exercises (1.20, 2.10) 

\c-d\>\c\~\d\ 

and 

\c-d\ = \d-c\>\d\-\c\ = -{|c| 



I 



Idl}. 



Thus, - {kl - \d\) < k - d\ < {kl - \d\} and the result follows from 
exercise 1.20(1). 

3 Only (iv) is not immediate. We have 

d(x,y) = \x—y\ = \x — z + z — y\ < \x — z\ + \z — y\ 
= d(x,z) + d(z,y). 

4 Suppose that r + s\/2 = t, where t is rational. Then, provided s ¥= 0, 

vi - £=. 

5 

But the right hand side of this equation is a rational number (why?) and 
so we have a contradiction. 

5 We are given that a = r + s\/2 satisfies the equation ax 2 + bx + c = 0. 
Hence 

a(r + sy/2) 2 +b(r + s\/2) + c = 

{ar 2 + 2as 2 + br + c} + {2a + b}s^2 = 0. 

Since a,b,c,s and r are rational, it follows that (2a + b)s = 0. Thus 

a(r - sy/2) 2 + b(r - sy/2) + c = {ar 2 + 2as 2 + br + c) 

-(2a + b)s\/2 = 

and so & = r—s\/2 is also a root of the equation. 

6 Suppose that m 2 = 3n 2 . Then m 2 is divisible by 3 and hence m is 
divisible by 3. (Try m = 2k + 1 or m = 3k + 2.) Hence m =■ 3k. But 
then 9& 2 = 3n 2 , i.e. 2k 2 = n 2 and so n is also divisible by 3. 

Suppose that m 3 = 2n 3 . Then m 3 is divisible by 2 and hence m is 
divisible by 2. (Try w = 2k + 1 .) Hence m = 2&. But then 8A: 3 = 2n\ 
i.e. 4£ 3 = « 3 and so n is also divisible by 2. 

• Exercise 2.10 

1 (i) False (ii) true (iii) true (iv) false (v) true. 

2 By exercise 1.20(1), || -jc|<5 if and only if -5 <£ -x<8. This 
last inequality is equivalent to 5 >x — £ > — 6 which is, in turn, 



Solutions to exercises (2.10, 2.13) 1 77 

equivalent to % + 5 >x > g -5. But ft - 5, % + 5) = 
{x:£-5<;c<£ + 6}. 

3 (i) (0, 1): bounded above; some upper bounds are 100, 2 and 1 , the 
smallest upper bound is 1 ; no maximum. 

(ii) (— °°, 2] : bounded above; some upper bounds are 100, 3 and 2; the 
smallest upper bound is 2 and this is also a maximum. 

(iii) {—1,0, 2, 5}: bounded above; some upper bounds are 100,6 and 
5; the smallest upper bound is 5 and this is also a maximum. 

(iv) (3, °°): unbounded above. 

(v) [0,1]: bounded above; some upper bounds are 100, 2 and 1 ; the 
smallest upper bound is 1 and this is also a maximum. 

4 (i) (0, 1): bounded below; some lower bounds are — 10, - 1, and 0; the 
largest lower bound is 0; no minimum. 

(ii) (— <*>, 2\. unbounded below. 

(iii) {—1,0,2, 5}: bounded below; some lower bounds are —10, —2 
and — 1 ; the largest lower bound is — 1 and this is also a minimum. 

(iv) (3, °°): bounded below; some lower bounds are — 1 0, 2 and 3 ; the 
largest lower bound is 3; no minimum. 

(v) [0, 1] : bounded below; some lower bounds are — 10,-1 and 0; 
largest lower bound is and this is also a minimum. 

5 {3,4}. 

6 Take y = \x. No m e (0, °°) can be a minimum because \m is a smaller 
element of the set. \ 

• Exercise 2.13 

1 Since B = sup 5 is an upper bound for S,x<B for each x G S. But 
S C5 means that, for each x G S , it is true that x^S. Hence x < B 
for each x G S . Thus B is an upper bound for S and therefore at least 
as large as the smallest upper bound sup S , i.e. sup S < B = sup S. 

2 The proof is very similar to that of theorem 2.12. Let B = sup S and let 
T = {x + % : x G S). Since x < B for any x G S, it is true that 

% + x < \ + B for any x G S and hence that % + B is an upper bound 
for T. If C is the smallest upper bound for T, it follows that C<%+B. 
On the other hand, y < C for any y G T and therefore y — | < C— % 



1 78 Solutions to exercises (2.13) 

for any y G T. Since S = {y — £ :y G 7"}, it follows that C — £ is an 
upper bound for S and therefore that B < C — £. 

We have shown that C< £ + B and O £ + 5. Hence C = £ + tf . 

3 Let B = sup 5 and let 7" = {— x : x G 5}. Since x < 5 for any x G 5 it is 
true that — x > —B for any x G S and hence that — B is a lower bound 
for T. If C is the largest lower bound for T, it follows that C > —B. 

On the other hand, y ~5* C for any y G 7" and therefore — j> < — C for 
any 7 G T. Since 5 = {— y :y&T), it follows that — C is an upper 
bound for S and therefore that —OB. 

We have shown that C > -B and C<—B. Hence C = -5. 

To obtain the required analogues, consider a non-empty set 7" which 
is bounded below and then apply theorem 2.12 and exercises 2.13(1) 
and (2) to the set S = {- x : x G T). We obtain 

(i) inf $x = | inf x (£>0) 

i£r x e t 

(ii) If J C 7\ then inf T >MT 
(iii) sup (— x) = — inf x. 

x&T xGT 



4 (i)l 



(ii)2 



(iii) 1 (iv) 0. 



5 (i) The set /J = {|| -x| :x G 5} has as a lower bound. If £ G 5, then 
is a minimum for D. Exercise 4(iv) provides an example for which 
tf(£,S) = Obutt£S. 

(ii) Let £ = sup S. Then |£ -x| = % -x for each x G 5. We therefore 
have to show that no ft > is a lower bound for the set 
D = {£— x :xGS}. If this is false, we can find an h > such that 
% -x > h for all x G S. But then x < % - h for all x G S and hence 
£ — ft is an upper bound for S smaller than the smallest upper bound. 
If £ = inf 5, consider instead d{~ I, T) where T= {-x :x G S}. 

(iii) Since / is an interval, | $ / implies that £ is either an upper bound 
or a lower bound for /. Suppose the former. Let B be the smallest upper 
bound of/. Then B G / because / is closed. Given any x G /, 

||— #| = %~x = S-B+B-x = $-B+\B-x\. 

Hence, by exercise 2.13(2), 

inf »— x| = £-fi+ inf \B-x\ 
x e s % * 4 s 

and therefore d(|, 5) = £ - £ + d(B, S). But £ - B > 0, t/(Z?, S) > 
and d(t, S) = 0. It follows that £ = 5 and hence £ G /. Similarly if £ is 
a lower bound for /. 



Solutions to exercises (2.13, 3.6) 



179 



If/ is an open interval other than R or then £ can be taken as its 
supremum or infimum (at least one of these must exist). 

6 We may assume that the sets S and Thave no elements in common 
(otherwise the problem is trivial). With reference to the given 'hint', the 
set To is not empty because t G T . The set T is bounded below by s. 
Let ft = inf TV 



■f 



I huh m il i 



* 



To 



If ft ^ T, then ft G S. But ft is at zero distance from T by exercise 
2.13(5ii) and hence we have found a point of S at zero distance from T. 
If ft G T, then ft > s and the interval (x, ft) is a non-empty subset of 5. 
Hence ft is a point of T at zero distance from S. 



• Exercise 3.6 
1 We have 

n n 

for all n G (^ and therefore 1 is an upper bound for S. Suppose that 
1 — ft, where ft > 0, is a smaller upper bound. Then, for all n G f^, 

l--<l-ft 

n 

n 

1 -\ 

' ?< ft 

and hence ft" 1 is an upper bound for N. This is a contradiction and 
therefore 1 is the smallest upper bound for S. 

The set S has no maximum. For no n G [^ is it true that 



h-1 



= 1. 



2 Suppose that S is bounded above. Then it has a smallest upper bound B. 
Since BX' 1 cannot be an upper bound, there exists an n G N such that 

X">BX' 1 
X"* l >B. 



180 Solutions to exercises (3.6) 

Hence B is not an upper bound for >*> and we have a contradiction. 

It is clear the is a lower bound for T. Suppose that h is a larger 
lower bound (i.e. h > 0). Then, for any « S fsj, 

x n >h 

l\" 1 

It follows that the set S = {(llx)":n G M} is bounded above by h' 1 
which contradicts the first part of the question. 

3 Let S be a non-empty set of integers which is bounded above by B. 
Since F*J is unbounded above, we can find an n G N such that n> B. 
The set T~ {n— x:x£ S} is a non-empty set of natural numbers and 
hence has a minimum m (theorem 3.5). The integer n — m is then the 
maximum of the set S. 

Similarly if S is bounded below. 

4 Since N is unbounded above, there exists a natural number n such that 
n>(b —a)" 1 . Otherwise (b —a)~' would be an upper bound for f^. Let 
m be the smallest integer satisfying m>an. Then m — 1 < an. Hence 



I 



m m 1 

a < — and — < a + — < a 

n n n 



(b-a) = b 



as required. 



5 Suppose thatm &S. Then \{tn + I)G5and^(m + \)>m. Hence m 
cannot be a maximum. Similarly m cannot be a minimum. 

It is obvious that S is bounded above by 1 . Suppose that H 
(0 < H< 1) were a smaller upper bound. By exercise 3.6(4) we could 
find a rational number r such that H<r<\. But then r would be an 
element of S larger than//. Thus 1 is the smallest upper bound. 

A similar argument shows that is the largest lower bound. 

6 We know that all numbers of the form r\j2 are irrational provided r is 
rational and non-zero (exercise 1 .20(4)). By exercise 3.16(4) we can 
find a rational number r such that 

and hence a < r\j2 < b. (If r happens to be zero, locate a second 
rational number s satisfying <s\J7 < b.) 



Solutions to exercises (3.11) 
Exercise 3.11 



181 



1 (i)Lets„= l 2 + 2 2 



'? + . . . + n 2 and let P(n) be the proposition that 
s„'= ««(« + 0(2n + 1)- Theni'(l) is true because s, = 1 and 
£l(l + 1)(2 + 1) = 1 . We next assume that P(n) is true and try and 
deduce thatP^i + 1) is true. 

s n + , = l 2 +2 2 +... + n 2 +(n+l) 2 = s n + (n+l) 2 

= \n(n + l)(2n + 1) + (n + l) 2 (assuming P(n) true) 

= K» + l){«(2n + 1) + 6(n + 1)} = l(n + 1)(2« 2 + In + 6) 

= l(n + 1)(« + 2)(2« + 3) = l 6 m(m + l)(2m + 1) 

(m=n + 1). 

(ii) Let s n = l 3 + 2 3 + . . . + n 3 and let P(n) be the proposition that 
s n = kn 2 (n+ l^.ThenPO^s true because s, = 1 andjl 2 (l + 1) 2 = 1. 
We next assume that P(n) is true and try and deduce that P(n + 1) is 
true. 

s n +i = l 3 + 2 3 +... + « 3 + («+l) 3 = s„ + («+l) 3 

= \n 2 (n + l) 2 + (n + 1 ) 3 (assuming P(n) true) 

= \(n + 1)V + 4n + 4) = |(« + \)\n + 2) 2 . 

2 Let P(n) be the assertion 



1 +x + ...+x" = 



1 -x 



Provided x ¥= 1 , this is clearly true when n = 0. Assuming P(n) is true, 



1 +x+x 2 + .. . + x n +x n * , = 



1 -x 



n + l 



1 -x 
l-x" + , +x" + 1 -x r 



\ 



l-x 



1 -x 



n + 2 



\-x 



and therefore P(n + 1) is true. 

The second formula of the question is obtained by writing a- =a/b. 
(Note that, if a = b or b = 0, the formula is obviously true.) 

3 Suppose that P(!-) = 0. Then 
P(x) = P(x)-PQ) 

= a n {x n -% n ) + a n . l {x"- i -% n - i )+ ...+ ai {x-%). 



182 Solutions to exercises (3.11) 

By the previous question we have 



x*-r = (*-£>(•* 



fe-l -L „&-2 



s 



l-fc-2 J. f-fe-1 



+ r - l) 



where F k _ t (x) is a polynomial of degree A: — 1 . It follows that 

/>(*) = (* -©WW*) +« B - 1 F n _ a €>c) + • • • + «,} = (a- -$)8(x). 

We apply this result to the second part of the question and obtain, in 
the first place, P(x) = (x — li)2i(x) where Qi(x) is of degree n — 1 . 
Then = P(% 2 ) = (£2 - f OGi(h). Since ? 2 # | b it follows that 
Qi{% 2 ) = and hence that Q t (x) = (x - % 2 )Q 2 (x) where Q 2 (*) is of 
degree n — 2. ThusP(x) = (x — £i)(x — % 2 )Q 2 (x). Next consider £ 3 and 
so on. 



B! 



■ + - 



«! 



r-lJ r!(n-r)! (r- !)!(« -r + 1)! 



H! 



(r-l)!(n-r)! Ir n-r+1 



£ 



1 



n\ 



n + 1 



(r-l)!(n-r)! r(n-r+ 1) 
fe + 1)1 /« + l\ 



r!(«-r+ 1)! 
Let P{n) be the assertion that 



(a + b)"= I \ \a- r b r . 



r=0 



Clearly P(l) is true. Assuming P(n) is true, 
(a+Z>)" +1 = (a + b)(a + b) n 



= a 



z 

(r=0 



a b 



+ b 



= £ f a »-v+ i(V~ r * r+I - 



= n\>- 



r = 



s:K'»' 



We replace r by s in the first sum and replace r + 1 by s in the 
second sum. Then 



(a + b) n * 1 = £ I U""' +1 6'+ I "J*""** 1 * 



Solutions to exercises (3.11, 4.6) 



183 



= a" 



z 



li 1 /« + 1 



= z 

s = 



+ , 

si \s-l 

n + 1 -slS 



_n+I -sts 



6* + 6" 



as required. 

5 Suppose there exists a natural number k for which P(k) is false. Since 
the set {2" : n e M} is unbounded above, we can find an M = 2 N > k. 
Now put S = {n : i>(") is false and n < M). This is a non-empty set of 
natural numbers which is bounded above. From exercise 3.6(3) it has a 
maximum element m. The assertion P(n) is true for m < n < M and 
hence P(m + 1) is true. But P(m + 1) implies P(m) and hence we have 
a contradiction. 



6 From exercise 3.11(2), 
a "- b " = (a-b)(a"- 1 + a n - 2 b + ...+ab"- 2 
Hence, if a >b >0, 
a" -6" 



fr"" 1 ). 



nb" 



< 



a-b 



<na" 



Therefore, 



If x > y, the result follows on writing a = x 1 '" and b =y 1/n . If 
.v<^,writea=>' ,/ "and6 =x Vn . 

• Exercise 4.6 

1 Let e > be given. We must find a value of A' such that, for any n >N, 

n 2 -\ 



n 2 +l 

But 

« 2 -l 



-1 



-1 



<e. 



= 1 - 



n 2 -\ 



n 2 +\ n 2 +\ 



n 2 + 1 

and so we simply need to find an N such that, for any n>N, 

,7TT <e - 

i.e. 



\ 



184 Solutions to exercises (4.6, 4.20) 

n 2 + 1 > 2/e. 

Choose N = (2/e) 1/2 . Then, for any n > N, 

n 2 + \>n 2 >N 2 = 2/e. 

We have shown that, given any e > 0, we can find an /V (namely 
(2/e) 1/2 ) such that, for any n >N, 






n 2 -\ 



n 2 +l 
Thus 

lim 



-1 



<e. 



n 2 -l 
n 2 +l 



= 1. 



2 Let e > be given. We must find a value of TV such that, for any n >N, 



7" Q 



<e 



i.e. 

« r >l/e. 

We choose N = (1 /e) 1/r . Then, for any « > TV, 

n r >N r = 1/e 

and the result follows. 

3 Let e > be given. We must find a value of N such that, for any n > N, 
\\x n -\l\<e. 

If X = there is nothing to prove. We may therefore assume X # 0. We 
are given that x n -* I as n -> °°. Since e/|X| > it follows that we can 
find an N such that, for any n>N, 

\x n ~l\<el\X\ 

i.e. 

|Xx„-X/|<e. 

This completes the proof. 

• Exercise 4.20 

n 3 + 5n 2 + 2 1 + 5/T 1 + 2n~ 3 1+0 + 



1 



- as n •* °°. 



2n s + 9 2 + 9n" 3 2 + 

2 First consider the case when |x|< 1 . Then x" •* as n -><». Hence 



Solutions to exercises (4.20 J 

x+x" x + 
1 +x"^ 1 +0 



185 



as n 



Next suppose that x > 1 . Then x " ■+ as n -*■ °°. Hence 
x+x" x.x~" + l + 1 



asn 



1 +x n x'"+l 0+1 
This leaves x = 1 and x = —\ giving 



1 + 1" 



= 1 and 



-1 +(-!)" 



1 + 1" * 1 +(-l) n ' 

The latter expression is not even defined for odd n. Hence 

1 (x> 1) 

x (— 1 < jc < 1 ) 

1 (*<-!). 



lim 



x+x" 



■»\l+x n 




limit 
undefined 



3 We have 



0<V(«+ i)-V« = 



(V(n+I)-V")(V(n + l)+Vn) 



(V(«+l)+V«) 



n+l-n 1 



as« -*-°° 



V(«+l) + \/n V" 
(see exercise 4.6(2)). It follows from the sandwich theorem that 

V(h + 1) — V" "* as n •* °°. 

4 Let N be the smallest natural number such thatN>x. Then, for n>N, 

x" X X X X X 

n~\ " T'2'N—l'N'"N 



X 



,v-i / \n-N+l 
X IX 



i! I 2 N-l N 

Since x/N < 1 , (x/N)" -* as « -* °° and the result follows from the 
sandwich theorem. 



186 Solutions to exercises (4.20, 4.29 "J 



5 Suppose that (1 + l/») a+1 1*| > 1 for all n e N. Then 



n 



\x\ 



for all « S PiJ and this is a contradiction (see example 3.4). Hence, for 
some We W, 

(1 +UN) a + 1 \x\<l. 



If x ¥= 0, consider the expression 
a+l 



(n+lf +1 x 



iOr-H„n + l 



n a+1 x n 



I 

n 



1*1. 



Un>N, 

\(n + l) a * l x n + 1 \<:\n a * 1 x n \. 

It follows that, for n>N, 

\n a+i x n \<\N a+1 x n \. 

We conclude that, for n>N, 



1/iVK-UV' 
« 



Q+l^jVl 



and thus n°x" ■* as n -» °° by the sandwich theorem. 
6 We wish to prove that 

This inequality is equivalent to 

(n + \) n <n n * 1 

i.e. 

But we know from example 4.19 that 



(i) 



i + 



<3 



for all n € H, Hence (1) holds provided 3 < n. 

• Exercise 4.29 

1 (i) Let e > be given. We have to find an N such that, for any n>N, 
||x„-/|-0|<e. 



Solutions to exercises (4.29) 



187 



But \\x n — /| — 0| = \x„ — 1\ and so what has to be proved is just 
the definition of x n -* / as n -* °°. 

(ii) By exercise 1.20(2), 

||.x n |-|/ll<U„-/| 

and hence \x n \ -*■ |/| as n ■* °° by the sandwich theorem. 

2 For any e > 0, there exists an N such that, for any n>N, 
l* B -l|<e. 

i.e. 

/-e<^:„</ + e. 

Since this is true for any e > 0, then it is true when e ■ "»|l (for an 
appropriate value of N, say W = A^). For « > TV, we therefore have that 

x n >l~\l - R 

3 (i) Let // > be given. We have to find an N such that, for any n>N, 

T>H. 

Since the set {2" : « G M} is unbounded above (exercise 3.6(2)), there 
exists an JV6 N such that 2 N >H. If n >N, then 

2">2 N >H 

and the result follows. (Alternatively one may appeal to the inequality 
2" > n {n > 1) which is easily proved by induction.) 

(ii) Let H > be given. We have to find an N such that, for any n>N, 

-s/n<-H 

i.e. y/n>H 

i.e. n>H 2 . 

We may therefore take N = H 2 . 

(iii) The sequence <(— l)"n) cannot converge because it is unbounded. 
It cannot diverge to + °° because its odd terms are all negative. It can- 
not diverge to —°° because its even terms are all positive. 

4 (i) Suppose that x n •* as n ■* °°. Let H > be given. We have to find 
an N such that, for any n >N, 

\/x n >H. 

But H' 1 > 0. Since x n -* as n -*■ °°, we can therefore find an N such 



1 88 Solu [ions to exercises (4. 29, 5. 7) 



that, for any n>N, 

x n = fx.-OKJT 1 

i.e. \lx n >H. 

(ii) To prove that \/x n -* + °° as n -* °° implies that x n -*■ as « -> °°, 
one simply has to reverse the above argument. 

5 Let H > be given. If Cx„) is unbounded above, there exists an N such 
that x N > #. If (x n ) increases, then, for any n >N,x„ >x N > H. It 
follows that x n -*■ + oo. Similarly in the case when 0c„) decreases and is 
bounded below. 

6 Suppose there is a value of n GN for which no x G S can be found 
satisfying |£ — x\< 1/h. Then 1/n is a lower bound for the set 

D = {\%—x\:xGS} which contradicts the assertion that dQ, S) = 0. 
That x„ -* i- as w -»■ °° follows from || — jc„| < l/« (« = 1 , 2, . . .) be- 
cause of the sandwich theorem. 

If £ = sup S, then from exercise 2.13(5ii) rf(|, 5) = 0. It therefore 
follows from the first part of the question that a sequence Cc n > of 
points of S can be found such thatJ£ n -» £ as « -» °°. (Note that the 
terms of this sequence are not necessarily distinct.) 

If S is unbounded above, then, given any n GH, we can find an 
x n G S such that* n > n. Hence x n -> + °° as n -> °°. 

• Exercise 5.7 

1 Suppose that « 1/n -> / as n ■* °°. It follows from theorem 5.2 that 
(2rt) 1/2n -<-/asH-°°. 

But 2" 2n ■* 1 as n -*■ °° (example 4.14). Thus n v2 " -> / as n -*■ °°. Hence 

It follows that / = P. Since / > 1 , we deduce that / = 1 . 

2 We prove that a<x n < bby induction. It is given that a < x 1 < b. If 
we assume that a < x n < b, it follows that 

x n+1 — a = x% + k—a>a 2 — a + k = 0. 

Similarly 

x n + s -b =xl+k-b<b 2 -b + k = 0. 

Hencea<x n+ , < b. (Note that the condition < k <$ ensures that 
the quadratic equation x 2 — x + k = has two real roots and that these 
are both positive.) 



Solutions to exercises (5. 7) 



189 



Next consider 

x n + l x n = x n X„ + k. 

But*?,— x„+k< because a <x n < b (see exercise 1.12(4)). This 
completes the proof that a < x n + 1 <x n < b. 

Since Cc„> decreases and is bounded below by a, it converges. 
Suppose that x n -* / as n -+ °°. Thenx n + 1 ->-/as« ^-°°. Butx n + i = 
x„ + k and so 

I = I 2 + k. 

This implies that l = aotl = b. But b cannot be the infimum of the 
sequence (x„> because it is not a lower bound. Hence / = a. 

3 It is easily shown by induction that k > x„ > (n = 1 , 2 , . . .). Observe 
that 



l n + l 



X n -1 — 



l+X n l+*„- 2 
k(X„-2 x n) 

Hence x n+i —x n . 1 has the opposite sign tox„.. 2 — x n . It follows, using 
an induction argument that one of the sequences <x 2n ) and <x 2n _j> 
increases and the other decreases. (In fact (x 2n -i) increases ifx 3 >*i 
and decreases if -X 3 < x t .) 

That both sequences converge follow from theorem 4.17. Suppose 
that x 2n •* I as n -*■ °° and that x 2n - 1 "* W as n -* °°. Then 

k 



1 = 



k 

1 +m' 



m 



l+l 



I + Im = k; m + Im = k. 

It follows that / = m and that I 2 + / = k as required. The conclusion 
about Ct,,) is that x n -*■ I as n -> °°. 

4 If < a < 6, it is easily seen that the geometric mean G = V(a&) (see 
example 3.10) and the harmonic mean//= {\(\ja + 1/6)}" 1 (see 
exercise 1.12(5)) satisfy 

a<H<G <b. 

f 

We are given h= x i<yi= 1. On the assumption that x n _, <^ n _ 1 , 

x n-l< x n<y n -l 

because x n is the geometric mean ofx n _, andj'n-!. Further 



190 Solutions to exercises (5. 7) 

x n <y n <y n -i 

because y n is the harmonic mean of x n and y n -j. It follows by induc- 
tion that x„_i <x n <y n <y„. l (" = 2, 3, . . .). 

The sequence (x n ) increases and is bounded above by y% — 1. The 
sequence (y n ) decreases and is bounded below by Xi = \. Hence both 
sequences converge. Suppose that x n -* I as n -* °° and y n -> m as n -> ° 
Then 

I 2 = Im 

m 2 \l m / 

Both of these equations yield I = m. 

5 As in example 4.19 we apply the inequality of the geometric and arith- 
metic means with ai — Ot — . . .— fl„-i = (1 +>'/(" — 0) an d a n = 1 • 
Then 

. (n-l)/n 
II 



y 

n-\ 



< (»-l)(l+K« -!)-')+! _ L + Z\ 
n \ n 



Hence 



£T4+i 



This analysis is only valid if a u a 2 , ■ ■ . , a n are non-negative numbers 
and hence we have only shown that the sequence increases when 



1 + 



y 



>0 i.e.n>\-y. 



n-l 
As in example 4.19, 

1+ 4" <1+w+ a + ... + fi£. 

nj 2! n\ 

Since there exists an N such that, for any n>N, 



ff<6 



(2) 



we may obtain from (2) 



iJV 



1 _/■»•»" + ! 

<C+ 33— <C+2 



where Cis a constant. 



r ♦■•■♦& 



Solutions to exercises (5. 7, 5.73^ 



i9/ 



It follows that -the sequence is convergent since it increases and is 
bounded above. Note that the limit is positive since 1 + y/n is positive 
when n is sufficiently large. 

6 From the previous question we know that the sequence 

x 2 
] ~n 2 
converges to a positive limit /. It follows that positive numbers a and b 
can be found such that 



for sufficiently large values of n. Hence 

«""<H"H" <4 '" 

The result then follows from the sandwich theorem and example 
4.14. 

• Exercise 5. 15 

1 For the sequence <(- 1)"(1 + l/«)> the set L = {- 1, 1}. Observe that 

(-l) 2n (l + l/2n)-> 1 as >?->•«• 
(_1) 2 " -1 (1 + l/(2«-l ))-*•-! as«-*-°° 

and the sequence possesses no subsequences which tend to a limit other 
than 1 or — 1 . It follows that 

(i) limsup(-l)"(l + l/«) = 1 00 lim inf (- 1)"(1 + l/«) = -1. 

Note that 
sup (-1)"(1 + 1/n) = (-1) 2 (1 +i) = I 

n > 1 

inf (-1)"(1 + 1/n) = (-l) l (l + 1/0 = -2. 

n > 1 

2 Observe that every rational number in the interval (0, 1) occurs infinite- 
ly often as a term in the given sequence. 

Let x £ [0, 1 ] . By exercise 3.6(4), a term /•„_ of the sequence can be 
found such that 

x-\<r„<x+ 1. 

A term r„, with n 2 >«i can then be found such that 



192 Solutions to exercises (5.151 

x-\<r n <x + \. 

Continuing in this way we construct a subsequence <r„ fc > such that 

and hence r„ fc -* x as k •* °° by the sandwich theorem. 

3 The hypothesis of the question implies that Ce n ) contains a subsequence 
all of whose terms are at least as large as b. By the Bolzano- Weierstrass 
theorem, this subsequence in turn contains a convergent subsequence 
<x„ r >. Suppose that x„ r -* I as r ■+ °°. Since x„ r > b (r = 1 , 2, . . .), it 
follows from theorem 4.23 that / > b. 

4 Suppose it is false that, for any e > 0, there exists an N such that, for 
any n >jV, x„ < /"+ e. Then for some e > it is true that for each A' 
we can find an n>N such that x n > 1 + e. From exercise 5.15(3) we 
can then deduce the existence of a convergent subsequence with limit 
/ > /"+ e. This contradicts the definition of /"(see §5.12). 

The corresponding result for Hs that for any e > 0, we can find an 
A' such that, for any n >N,x n >/ — e. 

5 Suppose that [ = 1 = 1. Let e > be given. By the previous question we 
can find an A'i such that, for any n>N t , 



Solutions to exercises (5.15, 5.21) 



193 



(3) 



x n < I + e. 

We can find an N 2 such that, for any n>N 2 , 

l-e<x n . (4) 

Take N = max {#,, N 2 }. Provided n > N, inequalities (3) and (4) are 
true simultaneously. Hence, for any n >N, 

I - e < x n < I + e 

i.e. |x„-/|<e. 

Thus x n -> / as n -* °°. 

That x n -*■ I as n -*■ °° implies I = 1 = 1 is trivial. 

To prove the last part of the question one only has to note that, if 
m is the limit of a convergent subsequence, then / < m < I. 

The sequence <M n ) decreases because we are taking the supremum at 
each stage of a smaller and smaller set. Any lower bound for Cc n > is also 
a lower bound for <Af„>. It follows from theorem 4.17 that <M„> con- 
verges. Suppose that M n -+M as n -*■ °°. 

Suppose that x n -*lasr-+°°. Then, for n T > n. 






x nr <M n 

and hence KM n by theorem 4.23. From theorem 4.23 again it follows 
that / < M . This is true for each / e L and so /"< M. 

From exercise 5.1 5(4) we know that, given any e > 0, we can find 
an n such that, for any k > n, 

x k < 1+ e. 

Thus /"+ e is an upper bound for the set {x k : k > n} and it follows that 

M<M n <l + e. 

Since this is true for any e> 0, M < /"(see example 1 .7). 
We conclude that M = / 

i.e. lim I sup x k \ = Urn sup x n . 

n->°°\h > n ) n -* « 

• Exercise 5.21 
1 If«>m, 

\x n —x m \ < \x n - X„-i I + \x n - 1 -x n . 2 \ + . . . + \x m+l -x m \ 

<,a n - 1 + a"' 7 + . .. + a m 



= a" 



1-a" 



1 -a 



< 



1-a 



Given e > 0, choose N so large that, for any m>N, 

_m 

<e (Example 4.12). 



1 -a 



The result then follows. 

Exercise 4.20(3) provides the example y n = \/n (n = 1 , 2, . . .). 

2 Note first that a < x n < b (n = 1 , 2, . . .). Hence 

a ^. x n + 1 b 
b x n a 

Next observe that 

x n + 2~ x n + l ~ X n + \X n —X n+i = JC n + i(X n — X n + i) 
X n + l 



\X n —X r 



\X n *2 X n + i \ — |A„ A n + n 

Xn+2 ' Xn+l 



< 



a + b 



\X n -Xn+ll 



It follows that 



194 Solutions to exercises (5.21) 

, \n-l 



|x n + 1 -*„|< 



a + b 



\X 2 — Xy\ 



and so we may proceed as in the previous question. 
i It x n +2 = %(X n + x n + i), 

X n + 2 <" 2~X n + i = X n+ ! + 2~X n = . . . = X2 + jXj. 

Put l = l(x2 + kxt). Then 

*n + l "*" 2*n = 2' 

•^n + l-^ = l(/-^ n ). 

Hence 

l*n + i-/l = \\x n -l\ =...= — |*i - /| ->0 as h -*■■». 

It follows that x n ->■ / as w -»• °°. 

'* -^n + 2 = i*n + l*n/ > 
Xn + 2 x n+l ~ X n + i X n = . . . = X 2 Xi. 

Put /={*2*.}" 3 - Then 

and it follows that x n -*■ I as n -* °°. 

4 Let <x n > be a sequence of points in [a, Z>]. Since [a, fcj is bounded, the 
sequence <*„> is bounded. By the Bolzano-Weierstrass theorem, the 
sequence therefore has a convergent subsequence (x n >. Suppose that 
Xn r -* / as r -* 00. Since a < *„ r < 6, it follows from theorem 4.23 that 
a < / < b. Hence (ae«> converges to a point of [a, 6] . 

5 Suppose that /is unbounded above. By exercise 4.29(6), we can find a 
sequence <x n ) of points of / such that x n -*- + » as « -»• °°. But no sub- 
sequence of such a sequence can converge and so this is a contradiction. 
Hence / is bounded above. Let b = sup /. By theorem 4.29(6) we can 
find a sequence 0e„> of points of/ such that x n •* b as n «* °°. yl// sub- 
sequences of tx„> also converge to b (theorem 5.2). Hence be I. 
Similarly for a = inf /. 

6 Let (x n ) be a sequence of distinct points of 5. Since (x n ) is bounded it 
contains a convergent subsequence (x n >. Suppose that x n ■* § as /■ ■* °°. 
Then £ is a cluster point of 5. 



Solutions to exercises (6.26 J 



195 



• Exercise 6.26 
1 We have 

3rc-2 
n(n + l)(« + 2) 

It follows that 
& 3«-2 



1 5 

= + 



fft «(« + !)(« + 2) 



n m + 1 n + 2 



= (-*+!-# 



+ (-i+/-l) 
+Hf+X-A 




2 A convergent sequence is bounded (theorem 4.25). It follows that, for 
some H, 

a n <Hb n (« = 1,2,...). 

Thus the convergence of S"= 1 &n implies that of S„ = , a„ by the 
comparison test. 

Since / > 0, from exercise 4.29(2), there exists an N such that, for 

any n > N, 
a n >\lb n 

and hence the convergence of 2™=i a n implies that of 2~=i ft n by the 
comparison test. 

(i) Take a n = — and b n = -. Since £ - diverges (theorem 6.5), it 

2« « n=l" 

00 j 

it follows that £ — diverges. 

n=l *" 



196 Solutions to exercises (6.26) 

1 1 °° 1 

00 Take a n = — and b n = -. Again V - diverges. 

m — i n n - j ln — \ 

2 1 °° 1 

(iii) Take a n = 2 and b n = -5. Since £ -5 converges (theorem 

s 2 

6.6), it follows that £ . converges. 
„=i 1+3 

3 We have 

2n n 

b n = a n+ , +.. . + a 2n = 2 a fc - J] <z fe ->- as h -* °°. 

Since the sequence (a„> decreases, 
b n >na 2n >0. 

It follows from the sandwich theorem that 2na 2n "* as n -> ■». 
Similarly (2« — l)a 2n ., -*• as n -* °°. 

4 The series diverges because its terms do not tend to zero (see example 
5.11). 



5 (i) We use the ratio test. 



'n*l 



= ((« + l)!) 2 (2n)! = (n + l) 2 1 

~ (2(/i + 1))! * («!) 2 (2« + 2)(2n + 1) "* 4 

The series therefore converges. 

(ii) Again the ratio test may be applied. 

a n + l _ ((n+l)l )\ n+l (2n)\ 1 5 

a n -^TTyy. 5 ^7?^ 4 as " "* °°- 

The series therefore diverges, 
(iii) We use the nth root test 



asn 



n 



n+ 1 



1 + — I ) -» — as n -*■ °o. 



The series therefore converges because e > 1 . 
(iv) Again the nth root test may be applied. 



n 



,n + 1 



4" 



Vn 



— as n -*• °°. 
e 



The series therefore diverges because e < 4. 



Solutions to exercises (6.26, 7.16) 



197 



(v) We have 

V0L±JW« = I : _<1 („-l,2,...) 

n «(V(« + i) + V«) "' 

and hence the series converges by the comparison test (see theorem 

6.6). 

(vi) This series converges by theorem 6.13. 



6 We have 

*-»*£-B + H + - + 



1 .+. ! 



4fl - 3 4n - 1 2n 



= 1 



+ 



I + i-i + i 
2 3 4 5 

1 1 



2 

2n 



I-... + — 
2« + 2 2« + 4 An 



= s 2n + - 



JL + _l_ + ...+i 1 . 

k + 1 « + 2 2«| 



Here s„ denotes the /ith partial sum of the given conditionally con- 
vergent series with sum s. Observe that 



2 3 2n 



= 2 



1 1 
- + -■ 

2 4 



-H- 



2k 



1 
2n 



From this identity it follows that 



1 



1 



n+ 1 n+2 
Hence 

f3n = S 2n + 2* 2n "* I s aS « 



+ il 



Exercise 7.16 



1 

5 - 




1 
1 
1 
1 
1 
1 
J 


T 


1 
1 
1 
I 

Y//////7\ 




1 2 



198 Solutions to exercises (7.1 6 J 



Each vertical line meets the graph in one and only one point. The 
range is {2, 5}. The image of the set [1,2] is {5}. 




(i) The equation does not define a function from R to itself. 

The value x = 2 is an example of an x for which no corresponding y 

exists. 

(ii) The equation does not define a function from [— 1, 1] to R. The 
value x = is an example of an x for which two corresponding values 
of y (namely + I and — 1) exist. 

(iii) The equation does define a function from [— 1, 1] to [0, 1]. For 
each x satisfying — 1 < x < 1 there is precisely one y with < y < 1 
satisfying |xj + |.v| = 1, i.e.y = 1 - \x\. 

3 We have 

/o<fr) - me* ~ \=& « ! Ttl 1 :3 (o<,<i) 



+ g(x) l+4x(l-;t) 



1 -x 



Ix 



go fix) = g(f(x)) = 4/(*)(l -/(*)) = 4 - 

\l +xj \l +x 

(0<x<\). 

The two formulae give different results when x = 1 and hence define 
different functions. 

In order that/" 1 : [0, 1] -* [0, 1] exist, it must be true that, for each 
y satisfying < y < 1 , the equation 

l-x 

has a unique solution x satisfying < x < 1 . This requirement is easily 
checked by solving the equation 

y +yx = 1 —x 

-y 



x = 



l+y 



Solutions to exercises (7.16) 

The inverse function therefore exists and 
1-v 



199 



r'o) = 



(0<y<\). 



l+y 

(Note that / = r X )- 
The equation 

y = 4x(l -x) 

does not have a unique solution x satisfying < x < 1 for each y 
satisfying < y < 1 . For example, 

44(1-1) = 4.|(1 -f) = I. 
Thusg" 1 does not exist. 

4 The function /has no inverse function. When^ = — 1 , for example, the 
equation^ = x 2 has no real solution at all. 

The function g has an inverse function. For each>- > there exists 
precisely one x > such that y = x 2 which we denote by y/y (see 
§1.9). Hence 

g' 1 (y) = y/y O>0). 

5 We have j' =g(x)if and only ifx =g~ l (y). 

(i)g- 1 o g(x) = £-%(*)) - r'OO = x (xGA) 
(ii)g o g'Hy) = g(g~ l (y)) = g(x) = y 0' 6 B). 
When g is as in question 4, these formulae reduce to 
yjx 2 = x (x>0) 

(\/y) 2 = y (y>o). 

6 (i) We apply exercise 2.13. Put 7= {f(x):x<= S}. Then 

sup {f{x) + c} = sup {y + c] 

iES y e T 

= c + sup y = c + sup f(x). 

(ii) Let 

H = sup fix); K = sup gix). 

x e s «6S 

Then, for all *e S, 
fix) + gix)<H + K 



200 



Solutions to exercises (7.16, 8.15) 



and hence H + K is an upper bound for the set 

{f(x)+g(x):x£S}. 

The result follows. 

To show that equality need not hold, take S = [0, 1] and let 
f(x) = x(0<x<l)andg(x) = 1 -x (0 <*< 1). Then 
f(x) + g(x) = 1 (0 < x < 1 ). Thus 

sup {/(*)+ #00} = 1 
X es 

but 

sup /(*) + sup #00 =1+1=2. 
x es iss 



• Exercise 8.15 

1 Since rational functions are continuous at every point at which they are 
defined (theorem 8.13), we obtain 



x 2 + 4 1+4 5 

W Ai»H " 1-4 3- 




, . , ^ 73 + 5x 42 + 9 0+5.0 + 9 

(ii) im t; = z 

v '*-o 3* 23 + 7 3.0 + 7 


9 

* 7 



/(*) = 



3-JC (jc> 1) 

1 (* = 1) 



7x (x<\). 



(0/00 + 2 as* 
5 > such that 

l/W-2Ke 



1 — . Given any e > 0, we must show how to find a 



provided that 1 — 6 < x < 1 . Since we are only concerned with values 
of* satisfying* < 1 , we can replace /(*) by 2x. The condition 
1/00 — 2| < e then becomes \2x - 2|< e, i.e. 1 - |e < x < 1 + §e. We 
therefore have lo find a 5 > such that 



he 



1 - \e < x < 1 

provided that 1 — 6 < x < 1 . The choice 6 = ie clearly suffices. 



Solutions to exercises (8. 15) 



201 



(ii)/00 ->■ 2 as x -* 1 +. Given any e > 0, we must show how to find a 
5 > such that 

1/00 -2|<e 

provided that 1 < x < 1 + 5. Since we are only concerned with values 
of* satisfying* > 1, we can replace /(*) by 3 — *. The condition 
1/00 -2|<e then becomes |3 -x — 2| <e, i.e. 1 -e<x<l + e. 
The choice 5 = e therefore suffices. 
It follows from proposition 8.4 that 

lim /OO 

X -» 1 

exists and is equal to 2. Note that/(l) = 1 and so the function is not 
continuous at the point 1 . 

3 lf**0, 

. . . l+2*+* 2 -l 
/OO = = 2 + x. 

Since in considering a limit as * ■+ we deliberately exclude considera- 
tion of the value of the function at * = 0, it follows that 

lim /OO = lim (2+*) = 2. 

* -» x -* 

4 Let e > be given. We must find a 5 > such that |* — %\ < e provided 
that < |* — 51 < 5- The choice 5 = e clearly suffices. 

Observe that 

l/&)-|KlS-*l 

and hence /(*) •* % as * -> J by the sandwich theorem (proposition 
8.14). 



5 If£>0,let0<Jr<$< Y and take % =y in exercise 3. 11 (6). If 
X < x < Y, 

L Y 1 ' n Y- i \x-^\<\x 1 "'-^ /n \<-X lln X- 1 \x-^ 
n n 

and the sandwich theorem applies. 

To prove that /(*) -* as * -*■ +, we must show that, given any 
e > 0, we can find a 6 > such that 

x Vn m |/n_ | <£ 

provided that < * < 5 . The choice 5 = e" suffices. 



202 Solutions to exercises (8.15, 8.20) 



6 For any e > 0, we can find a 6 > such that 

|/(*)-/|<e 

provided < \x — £| < 5 . Since this is true for any e > 0, it is therefore 
true for e = I (with an appropriate value of 6, say 6 = h). We then 
obtain 

\f(x)-l\<l 

i.e.O = l-Kf(x)<l + l 

provided that < \x - §| < ft, i.e. §— h <x < £ + h and* ^ £. 

(i) if / < 0, an A > can be found such that /"(x) < provided that 
Z-h<x<H+handx*t. 

(ii) If / = 0, nothing of this sort can be said. 

• Exercise 8.20 

1 (i) Given // > 0, we must find a 5 > such that 

x' x <-H (5) 

provided that -6 <x<0. If x <0, (5) is equivalent to 1 >-Hx 
which is in turn equivalent to 

-H' l <x 

and hence the choice 6 =H~ 1 will suffice. 

(ii) Given H > 0, we must find an X such that 

x 2 >H 

provided that x > X. The choice A" = yjH suffices. 

2 Let e > be given. We must find an X such that 
|/fe(*))-/|<e 

provided that x > X. Since /(y) -* I as >> -* + °°, we can find a Y such 
that 

l/0)-/|<e (6) 

provided that 

y > Y. (7) 



Since g(x) 
x>X, 

g(x) > Y. 



1 as x -* + °°, we can find an X such that, for any 



Solutions to exercises (8.20, 9.1 7) 



203 



If x > X, it follows that (7) is satisfied with y = g(x) and hence that 
(6) is satisfied with;- = g(x). But this is what we had to prove. 

The problems of theorem 8.17 arise because of the possibility that 
g(x) = r? for some values of x. In the above problem 77 is replaced by 
+ °° which is not a possible value for g. 

3 Let e > be given. We have to find an X such that, for any x > X, 

Since /(y) -> / as >> -*■ +, we can find a 5 > such that 

l/0)-/|<e 

provided that < y < 5. 

We choose X = 5~ l . Then, if x > X, 

y = X H <X~* = 5 

and the result follows. 

4 Suppose that f(x) ■* \ as x -* £. Then f(x n ) -*■ X as n -> °° and f(y n ) -*■ \ 
as n -» °° (theorem 8.9). This is a contradiction. 

5 Takex n = 1/n (« = 1, 2, . . .) and;' n =y/2/n (n = 1, 2, . . .). Then 
/(x n ) = 1 -*■ 1 as n -» °° and /(y„) = -» as n -> «». Hence 

lim fix) 

x -* 

does not exist by the previous question. 

6 We have < \xf(x)\ < \x\ and so the result follows from the sandwich 
theorem. 



• Exercise 9.17 



1 





113 


— 2 
2j + 3 


k. ! i > 




J- = |.v - II 



Continuous on ( — 2, 2) and on 
[0, 1], Recall that a rational 
function is continuous wherever it 
is defined. 



Continuous on (—2, 2) and on 
[0, I ]. Note that the function has 
a 'corner' at x — 1 . 



204 Solutions to exercises (9. 1 7) 



Solutions to exercises (9.17, 10.11) 



205 



,l " 1 A> , 


1 - 




1 







[l * 



(iv) 



Not continuous on (— 2, 2) 
and not continuous on [0, 1]. We 
have/tr) -* 1 as X — ■ 1 — but 
/(l) = 2. 



(v) 



Not continuous on (—2, 2) but 
continuous on [0,1]. 




j"-*±Jer#±a 



Continuous on (— 2, 2) and on 
[0,1]. 



Continuous on (— 2, 2) and on 
[0,1], 



All of the functions except (v) are bounded on (—2, 2): all but (i) 
and (ii) attain a maximum on (—2, 2): all but (i) and (v) attain a mini- 
mum on (—2, 2). 

2 Let g: 1 -* R be defined by g(x) = f{x) — x 2 . Then g is continuous on /. 
Given any point g E I, let </•„> be a sequence of rational numbers in / 
such that r„ -* g as n -» °°. Then #(r„) ■* g(g) as n ■* °° (proposition 9.6). 
But g(r„) = (n = 1, 2, . . .). It follows that 

=*«) = /(g)-g 2 . 

3 Let P be a polynomial of odd degree n. Let 
/•(jc) = a n x n +a n . l x n - l + ...+a 1 x+a 

a„ + + . . . + -=rr. + -= 

x x ' x 

lfa n >0, then P(x) -* + °° as x -* + °°andP(x)^ — °°asx -*■— °°. 






Hence we can find a and b s©h that P(a) < < P(b). Since P is con- 
tinuous everywhere, it follows from corollary 9.10 that there exists a g 
between a and b such that P(£) = 0. (Similarly if a n < 0.) 

4 Suppose that g can be found with /(g) > 0. Since fix) -* as x -» + °°, 
we can find 6 > g such that, for any x > b, \f{x)\ < /(g). Also, since 
f(x) -*■ as x -* — °°, we can find a < g such that, for any x < a, 
\f(x)\ </(g). By theorem 9.12, /attains a maximum on [a, b]. 
Suppose that /(t?) > f(x) (x E [a, b ] ). Then /(t?) > /(g) > f(x) 

(x ^ [a, b]). Hence/attains a maximum on R. 

If a | can be found for which /(g) < 0, then a similar argument 
shows the existence of a minimum on R. 

If f(x) = for all x, the result is trivial. 

5 Let Xj E I. Then there exists an x 2 E I such that |/(x 2 )l < |l/C*j)l and 
there exists an x 3 E I such that |/(x 3 )| < \\f(x 7 )\ and so on. We can 
therefore find a sequence (x n ) of points of / such that 

l/(*»)l < \ l/(*»-.)l < p 1/(^-2)1 < • • • < ^ l/fedl. 

It follows that /(x„) •* as m -* °°. Since / is compact, the sequence (x n ) 
contains a subsequence <je„ r > which converges to a point g E I (exercise 
5.21(4)). Because /is continuous on /,/(*„,.)-* /(g) as r -* °° (prop- 
osition 9.6). Hence /(g) = 0. 

6 That /is continuous on / follows from proposition 9.3. We have 
\x n + 1 -x n \ = \f(x n )-f(x n .i)\<a\x n -x n . 1 \. 

Hence 

|x B+1 -ayKaP-Haa -*il (« = 2, 3, . . .). 

As in exercise 5.21(1), we may conclude that (x n > is a Cauchy sequence 
and hence that 0c n > converges. Suppose that x n -*■ I as n -* ■». Because / 
is closed we have that / E I. But /is continuous on / and therefore we 
may deduce from the equation 



x n 


►1 = f{Xn) 




that /=/(/)• 




• Exercise 10.11 




1 We have to consider 




1 


1 ) 


1 


h 


1 + (x + hf \+x 2 \ 


h 



1 +x 2 -\ -x 2 -2xh-h 2 



(1 + (x + h) 2 )(l +x 2 ) 



206 Solu tions to exercises (10.11) 
-2x-h 



-2x 



(i + (x + hy)(i +x 2 ) (1+ x y 



as h -> 0. 



2 Let n = —m where m£N. Then, by theorem 10.9(iii), 
1 x m . - 1 . w*" 



Z). v " = D — = 



3 lfft>0, 
/(1+/Q-/(1) _ 2 + 2/; -2 



h 

lfh<0, 
f{\+h)-f{\) _ (1 +/!)'+ 1-2 



= 2 -► 2 as A ->■ ■ 



ft 



ft 



= 2 + ft->-2asft->0-. 



That/'(1) = 2 then follows from proposition 8.4. 
On the other hand, if h > 0, 

g(0 + h)-g(0) m ft-0 
ft ft 

but, if ft <0, 

*(0 + ft)-*(0) -A-0 



-> 1 as ft -* + 



ft 



ft 



• — 1 as ft -> — . 



4 By exercise 3.1 1(3), Pfr) = (x — g)R(x) where R is a polynomial of 
degree n — 1 . Thus 

/•'(*) = £(*) + (jc - %)R \x) (theorem 10.9(h)). 

Since P\%) = it follows that R (£) = and therefore /?(*) = 
(•* — Dfi(*) where Q is a polynomial of degree n — 2. 

5 Put #,(*) = (*-?)'. Then 

flfeft) = /(/-I)... (/-ft + 1)(* -©'"" (ft = 0,1, 2,. ..,/)■ 

It follows that £*&(£) = (ft = 0, 2, ...,/- 1) and 

D l g,(x) = /! (xGR). 

Hence D k g,(x) = (ft = I + 1, / + 2, . . .). We now apply these results 
to the 'Taylor polynomial' P and obtain 

D k p$) = o + o + . . . + i /o^dz^cd + . . . + o 

= / <fc, <£) (ft = 0, !,...,« -1). 



Solutions to exercises (10.11. 10.15) 207 

6 We prove the result by induction. It certainly holds for n— 1. If 

DJg = i h rffir-ig 

then 

D n + l fg = I ("] {D i+1 fD n - i g + D i fD n -^ 1 g} 

and the proof continues as in the solution to exercise 3.1 1(4). 

• Exercise 10.15 

1 Let.y >x. Then, by exercise 3.1 1(6) withX = * and Y=y, 

i v I/n_ y l/n 1 

It follows from the sandwich tlieorem that 

lim * — = -x^x~\ 

y-*x+ y— x n 

A similar argument shows that the left hand limit is the same. 

2 Write r = m/n where m is an integer and n a natural number. Then 
D{x m,n } = D{(x m ) vn } = - (x m ) 1 l n x- m mx m - 1 - -x m/B *- 1 . 

3 (i)Z){l +x" 27 } 3/5 = |{1 +x 1/2 T 2/S •^^" 26,27 (*>°)" 
(ii)5{x + (x+x 1,2 ) 1/2 } 1/2 

= I {x + (x + x v Y 2 }- V2 { 1 + le* + *" 2 )~ 1/2 (i + K" 2 )} 6» > °)- 

4 f {/(* 2 )} = /V)2x; f {/(*)} 2 = 2f(x)f\x). 
dx ax 

When x = 1 , 

2/'(l) = 2/(l)/'(l) 

and the result follows. 

5 By theorem 10.13, 

1 = Dx = /)(/-' o /)(*) = Df-^Dfix). 



208 Solutions to exercises (10.15, 11.8) 

y 




•v =£-'(>•) 



Observe that, for h > 0, 
(0 + /0 i/3_ i/3 = ^^ 

H 

Thus^" 1 is not differentiable at 0. 



°° as h -*■ +. 



6 If* is rational, then y =f(x) = x is rational and so 
fo f{x) = /(/(*)) = /O) = j, = /(*) = *. 
If x is irrational, then^ = f(x) = —x is irrational and so 
fO f(x) = f(fix)) = /O) = -y = -fix) = *. 

A thoughtless application of theorem 10.13 would yield 
1 



Dfix) = 



Dfiy) 



(V =/(*))• 



But the function /is not differentiable at any point - thus nothing can 
be deduced from theorem 10.13. 

• Exercise 11.8 

1 The stationary points are the roots of the equation /'(x) = 0. But 
fix) =x 3 -3x 2 + 2x and so 

fix) = 3x 2 -6x + 2 



'y=x(x- l)(x-2) 




Solutions to exercises (11.8) 



209 



and hence the stationary points are given by 
3±V(9-6) 1 



x = 



3 ^VS' 



If the maximum on the compact interval [0, 3] is not attained at an 
endpoint, then it must be achieved at a stationary point (theorem 
1 1.2). The only possible candidates for a point at which a maximum is 
achieved are therefore x = 0,x = 1 ± 3" 1 ' 2 and* = 3. If /is evaluated 
at each of these points, one finds that/(3) = 6 gives the largest value 
and hence 6 is the maximum. A similar argument shows that — 2/3 \/3 
is the minimum and this is achieved at x = 1 + 3~ 

2 Choose the constant h so that F = f + hg satisfies F(ff) = Fib) and thus 
Rolle's theorem may be applied. We need 

fia) + hgia) = fib) + hgib) 

i.e. 

fib) -fid) 

gib)-gia)' 
We obtain the existence of a £ G (a, b) such that 

o = f\%) = fm+hg'm 

i.e. 

g'm 

and the result follows. 



h = -' 



3 Apply the previous result with b = x. We deduce the existence of a 
% G (a, x) such that 

fa) m /(■*) 
g'm g&) ' 

Observe that % depends on the value of x, i.e. '£ is a function of x\ By 
exercise 8.1 5(4), we have % -*■ a as x -*■ a. Moreover %i= a when x > a 
and so it follows from theorem 8.17(ii), that 

,. fi%) .. f(y) 

hm -7— = hm ,, . 
*-* a+ gii) y-*°*g(y) 

provided that the second limit exists. The result follows. 



lim ^~V(1 +y 2 ) = lim . 

y -►too"' * V " ' X->0+ \X 



'' + ? 



21 Solu tions to exercises (11.8) 



= lim 



1 -V(l +x 2 ) 



= lim = 0. 

x -* o+ 1 



4 Given that /'(*) = x 2 for all x, it follows that 

D{f(x)-\x 3 } = f'(x)-x 2 = 

for all x and hence there exists a constant c such that fix) — %x 3 = c 
for all x (theorem 1 1 .7). 

5 From Rolle's theorem we may deduce the existence of numbers 

J?o.n»i • • • ,V n -i such that | <T?o<^i<'7i<- ■ -<g n -i <*?„-, <g n 
and 

f\Vi)-P\m) = (i = 0, l,...,n-l). 

Containing in this way we can demonstrate the existence of a £ such 
that 

f M i%)-P {n \%) = 0. 

But a polynomial of degree « — 1 has the property that P^"\x) = 
for all*. 

6 Since g(0) =g(i) = 0, it follows that/(0) =/(l) = 1 and so Rolle's 
theorem may be applied on the interval [0, 1]. We obtain the existence 
of a g £ (0, 1) such that /'(g) = 0. But then /(g) = 1 and so Rolle's 
theorem may be applied on the intervals [0, g] and [g, 1 ] and so on. 
This argument shows that between every pair of points x and .y for 
which f(x) — f(y) = 1 , we can find a point z for which /(z) = 1 . 

Suppose that < a < b < 1 and (a, b) contains no point x for which 
f{x) = 1 . Let 

A = sup{x:*<aand/(.x) = \}\B = inf {y :y>b and/O) = l}. 

Because a<b,A<B. Also, the continuity of /implies that 
fiA) =fiB) = 1 [use exercise 4.29(6) and theorem 8.9]. But we know 
that a point C can be found between A and B for which /(C) = 1 and 
this is a contradiction. 

Let* G [0, 1]. Since every open subinterval of [0, 1] contains a 
point g for which /(g) = 1, we can find a sequence g„ of points of 
[0, 1] with/(g n ) = 1 such that % n -*x as«^-°°. This implies that 
fix) = 1 because /is continuous. 



1 



Solutions to exercises (11.11) 



211 






• Exercise 11.11 

1 Let /: U -> U be defined by fix) = x" and apply Taylor's theorem in 
the form 

f(y) = m)+Y^-^'^ + --- + h (y ' m ' n)m+En+v 

Put y = x + 1 and g = 1 . One then has to observe that, for 
/c = 0,l,.. .,«, 

n(n -!)■■■(« -* +l)g"- fe i 
ft! 



j^/ni) = 



and that 



1 



X n + 1 / (n + l) (7?) = 0. 



(" + 0! 
2 By Taylor's theorem 

0+4) 1 ' 2 = 2+^4-" 2 +^-Ki-i)4- 3 ' 2 +j ? -ia-i)a-2)'?- s/2 

where 17 lies between 4 and 5. Hence 
But 



n 



5/2 -v, 45/2 _ 2 s 



and the result follows. 
3 By Taylor's theorem, for any xGl, 

fix) = /(g) + ^A?) + .-. + ^9"/ (,, - 1) «) + ^ 

Hence, with the given information, 

for some 7? between * and g. Since / (n> is continuous at g, it follows 
from exercise 8.15(6), that, if x is taken sufficiently close to g, then 
/ (n) (7?) will have the same sign as/ (n, (g) (provided that/ (n) (g) ¥> 0). 
The problem therefore reduces to examining the sign of the right 
hand side of (8) under the various cases listed. 

(i)/ (n >(g) > and n even. Then the right hand side of (8) is positive for 
x sufficiently close to g and hence /has a local minimum at g. 



(8) 



212 Solutions to exercises (11.11, 12.12) 

(ii)/ (,1> (£) > and « odd. If x is sufficiently close to f, then the right 
hand side of (8) is positive if x > £ and negative if x < %. 

(iii)/*"^!) < and n even. We obtain a local maximum. 

(iv)/ (n) (?) < and n odd. This case is similar to (ii) - see the diagram 
below. 



(i) 




Solutions to exercises (12.12) 



213 



(iii) 



(iv) 







Of the three functions given, (v) falls into class (ii), (vi) falls into 
class (i) and (vii) into class (iii). 

• Exercise 12.12 

1 We have f(a) < f(x) < f(b) (a<x< b). 

2 Observe that 

/'(*) = -(_x+]) ll "(x+\)- , --x l "'x- 1 <0 (x>0) 
n n 

because n > 1 and hence 

i+ £)H) >] (x>oy 

3 If x and y lie in /and. y >x, then f(y) >f{x). Hence 

m= Wm m=M m lim /W^,o. 

y->* y— x y-**+ j — x 

The function/: R ■+ R defined by /(a:) = x 3 is strictly increasing on 
Rbut/'(0) = 0. 

4 Since/is continuous, /(R) is an interval (theorem 9.9). Since f(x) -* 
+ °° as x -» +«> and /(x) ■* - °° as x -*■ — °°, it follows that /(R) = R . 



We have 

fix) = 1 + 3x 2 >0 (*eR) 

and hence theorem 12.10 assures us of the existence of an inverse func- 
tion f' 1 : R •* R . 

lie unique value of x satisfying — \=y = f(x) = 1 + x + x is 
obviously x = — 1 . Hence 

1 1 I 

Df(x) " £>/(-!) 4" 



DrH- 1) = Df-'iy) = 



5 Suppose that % is not an endpoint of/. To show that /is continuous at 
£, we need to prove that/(|-) =/(?+) (see corollary 12.5). 

If /(£ -) < /(£ +) then we can find X such that /(£ -) < X </(£ +) 
and X =#/(!). Choose a and ft in / so that a < % < b. Then /(a) < X < 
f(b) and so we can find an x e / such that /(*) = X and hence 

f(X ~) < /(*) < /$ +)• Since f&) *f(%) this is in consistent with 
corollary 12.5 and hence /(£-) =/(£ +). 

A slightly different argument is required if % is an endpoint. 

The function /: (0, 2) ■* R defined by 



m = 



x (0<x<l) 
X-J (Kx<2) 



shows that the requirement that /be increasing cannot be abandoned 
(although it might be replaced by some new condition). 

For each x satisfying < X < 1 , < f(x) < Mix). It follows that 
/(*)' 



0(x) = lim 



Mix) 



exists for each* satisfying <x < 1 and that 0(x) = or 0(x) = 1. 

If /increases on [0,1], then/(x) = Mix) for each x satisfying 
< x < 1 and hence 0(x) = 1 (0 <x < 1). Thus is continuous. 

It is more difficult to deduce that /is increasing given that is con- 
tinuous. We show to begin with that, if is continuous on [0, 1], then 

<fi(x)= 1 (G«X<1). 

We know that 0([O, 1 ]) C {0, 1}. But, by theorem 9.9, the image of 
an interval under a continuous function is another interval. Hence either 
0([O, 1]) = {0} or 0([O, 1]) = {1}. But the former case is impossible 
because 0(0) = 1 . Thus 0(x) = 1 (0 < x < 1 ). 

Let < x <y < 1 . We know that fix) = Mix) and fiy) = M(y). 
Obviously Miy)>Mix) and it follows that / increases on [0,1]. 



214 Solutions to exercises (12.21) 

• Exercise 12.21 

1 We have f(x) = (x - l)(x - 2)(x - 3) = x 3 - 6x 2 + 1 lx - 6. Hence 

f'(x) = 3x 2 -12x + ll 

f"(x) = 6x-12 = 6(x-2). 

If follows immediately that / is convex for x > 2 and concave for x < 2. 
The roots of the quadratic equation 3.x 2 — 12x + 1 1 are 

x = 2± 3" 1/2 

and it follows that /decreases on the interval [2 - 3~ l/2 , 2 + 3" 1 ' 2 ] 



' 



and increases on the intervals (— °°, 2 — 3 1/2 ] and [2 + 3 



-1/2 



")■ 




2 Take*i = x,x 3 =y andx 2 = ax + /3y in (1) of §12.13. Since 
a + 0=l, 

x 2 = ax t +(1 -a)x 3 

and 



a = 



x 3 -x 2 



= 



x 2 ~Xx 



X 3 Xi X 3 X\ 

Thus (1) of § 12.13 is equivalent to 

(*3 -*l)/(*2) < (*3 -X 2 )f(X 1 ) + (X 2 -X l )f(x 3 ). 

Consider now the inequality 

/(* 2 )-/T*i) ^ /(*3)-/(*i) 

X 2 ~Xi X 3 ~X t 

This is equivalent to 

(*3 ~xi)f{x 2 ) - (x 3 ~Xi)f{x{) < (x 2 -x,)/(*3) - (x 2 -x,)/^,) 

i.e. 

(x 3 -x 1 )f(x 2 )<(x 3 -x 1 -x 2 +x i )f(x 1 ) + (x 2 -x 1 )f(x 3 ) 



(9) 



Solutions to exercises (12.21) 215 

which is (9). A similar argument establishes the inequality 

/(*3) -/(*■) ^ /(*3)-/(*2) 
x 3 — x, x 3 —x 2 

3 LetZG7and Y&J. Set X = f(x) and Y=f(y). From (1) of §12.13, 
/(ax + ft) < a/(x) + 0/0) = ctX + H Y 

provided that a > 0, (3 > and a + (3 = 1 . Since /is strictly increasing 
on /, /"' is strictly increasing on J. Thus 

ar'(X) + $r\Y) = Qtx+$y<f~ i {aX+m 

and hence /"' in concave on J. 

If /is strictly decreasing on /, then/" 1 is strictly decreasing on J. In 
this case the last step in the proof is replaced by 

af\X) + PrXY) = ax+Py>r\aX + 0Y) 

and so/" 1 is convex on /. 

4 By the mean value theorem, there exists an t] between x and £ for 
which 



/(*)-/«) _ 



x-% 



= /'(*?)• 



Because /is convex, its derivative increases. Thus/'O?) >f (£) '€x > % 
and/'(r?) </'(£) if* < I Hence 

Geometrically the result asserts that the graph of the function lies 
above any tangent drawn to the graph. 



i-=/U) 




= /«)+/'«) (-v-S) 




/ concave 



If/ is concave the inequality should be reversed. 

The geometry of the Newton- Raphson process is indicated below. 



216 Solutions to exercises (12.21) 



<M*b)<I 




Given that x„ > %, <t>(x n ) > and <t>'(x„) > 0. (We cannot have <p'(x) = 
for any x because 0' is increasing and hence we would have <t>'(y) = 
(y < x)). Thusx n+1 <x n . On the other hand, by the inequality of the 
first part of the question, 

*(*n + l) -0(*n) >*'(**)(*,. + l-*n) = -0(*n) 

and hence (p(x„ + 1 )>0. It follows that £<x n+ , <x„. 

Since <x„> is decreasing and bounded below, it converges. Say 
x n -*■ I as n -*■ °°. Then <p(x n ) ■* 0(/) as n -* °° because is continuous 
and (p'(x n ) -*• <t>'(l +) as n -*■ °° because 0' is increasing. We obtain 



1 l *'(/+) 

and hence 0(/) = 0. Thus I = £. 

5 Suppose that /'(I) > 0. Then, using question 4, 
/(*) > /«) + f'(Mm ~ D - + °° as x ■* + «.. 
Similarly, if /'(£)< 0, 

/(*) >/KS) + /'©(•* -£>-* +«»as* -*- ~. 

It follows that /"(£) = for all values of % and therefore /is con- 
stant (theorem 1 1.7). 

6 Let P(n) be the assertion that 



/ 



Xi+X 2 + . . .+x n 



<-{/(*.) + ...+/(*„)} 



(10) 



for any x, , x 2 , ■ ■ . , x n in the interval /. That P(2) holds is given. We 
assume that P(2") holds and seek to establish P(2" + l ). Write m = 2". 
Then2w = 2" + 1 and 



^T^i-4 



-(x,+ .. . +x m ) + -(x m + 1 + . . .+x 2m )\ 






Solutions to exercises (12.21, 13.18) 

1 



217 



1 >■ + ••• +* 
■2 f [ 



* +f/ 



■ • • + x, 



< 2m {fXXl) + " ' + /( * m)} + 2m {f(Xm + t) + ' ' ' + /( * 2 '" )} 

- ~ {/0f0 + • • • + /(*am)}- 

2m 

We now assume that P(n) holds and seek to deduce that P{n — 1) holds. 
Write 



X = 



x t +x 2 + . . -+* n -l 

K-l 



Then 

/ jc 1 +x 2 + ■ ..+X n - i 
; \ n-\ 



= f(X)=f 



l(n-l)X + X 



X\ + X 2 + . . . + X n _] + X 



= / 



<-{/X*l) + ---+/(*n-l)+/W}. 



Hence 



I — -) /(*) <- {/(xi) + /(* 2 ) + ■ • • + /(*„ - 0} 

and P(n — 1 ) follows. 

Let a be a rational number satisfying < a < 1 . We may write 
a = m/n . If & = 1 — a, then = (/; — w)/« . Apply inequality ( 1 0) with 
x,=x 2 = . . .=x m = xandx m + 1 =x m + 2 = . . . =x n = y. Then 

/(ax +0>O <«/(*) + 0/00. (11) 

If a is irrational, consider a sequence <a„> of rational numbers such that 
a n -*■ a as n -*■ °°. Then n = I — a n -» 1 — a = as n -> °°. We have 

f(a n x + n y) < a n f(x) + n f(y) (w = 1 , 2, . . .)• 

Since /is continuous, consideration of the limit shows that (1 1) holds 
even when a is irrational. 



• Exercise 13.18 

1 If «S H, we have for any x 

1 



/J 



= X" 



n + 1 
and hence the result follows from theorem 13.14. 



218 Solutions to exercises (13.18, 13.26) 



If n is replaced by a rational number »#— 1, the same argument 
applies. The restriction < a < b is required because x a need not be 
defined for* <0 (e.g. if a = — 2,;c 0: is not defined for* = 0). 

If a = — 1, the argument fails. (Why?) 

2 The integrand is not defined when x = 1 and, in fact, has an unpleasant 
'singularity' at this point. 

3 As in example 13.17, we obtain 
^ n {0 a +l Q + ... + («-lf} 

<j 1 o x«dx<-^- 1 {\" + 2 a + . .. + ««}. 
It follows that 

and the result follows. 

• Exercise 13.26 

1 Suppose that, for some % G [a, b],g(%) > 0. Then, because # is con- 
tinuous on [a, b] . a subinterval [c, d\ containing £ can be found for 
which 

f(*)>k(D (c<x<d) 

(proof?). But then 



j a g(t) dt > j g g(t) dt > \g(£)(d - c) > 0. 



2 We integrate by parts. Then 
J b a xf"(x)dx = [xf(pe))*-f*f'(x)dx 

and the result follows. 

3 Observe that F'(t)=f(t) > {F(t)} xn (theorem 13.12). Hence 
(x - 1) = f 1 dt < f {F(t)}' U2 F\t) dt (theorem 13.23) 



Solutions to exercises (13.26) 



219 



= {2{F{t)Y n ] x i (theorem 13.14) 
= 2{F(x)y 2 <2f(x). 

4 By the continuity property (theorem 9.12), the function /attains a 
maximum M and a minimum m on the set [a, b]. Hence, by theorem 
13.23, 



I j^ g(t) dt < J o " f(t)g(t) dt<M j a " g(t) dt. 



(12) 



We now appeal to the intermediate value theorem (corollary 9.10). The 
function F defined on [a, b] by 

F(x) = /(*) f g(t)dt 

Ja 

is continuous on [a, b] and hence a % S [a, b] can be found such that 
F{%) = f t\t)g(f)dt 

J a 

because of (12). 

Ug(t) - 1 (a < t < b) and Fis a primitive for /on [a, b], then the 
result reduces to 

F(b)-F(a) = F'(£)(b-a) 

which is the mean value theorem (i.e. theorem 1 1.6) for the function F. 

5 We have 

j a b f(t) g (t) dt = [f(0G(t)) b a - jl noG(t) dt. 

Since f\t) > (a < / < b) we may apply the previous question and 
deduce the existence of | E [a, b] such that 



j" a t\t)g{t)dt = [/(/)G(f)L 6 -G«)J" o V(0 



df 



= [f(0G(t)] b a -G(k)im] b a 



= /(«) Jj«05 ^ + /(*) j { %« *'• 



6 We integrate the given integral by parts. Then 



220 Solutions to exercises (13.26, 13.34) 

■ _ t \n-lf(n-W f - i ,x 



[(x-ty-v^Xm 

+ J* (n-\)(x-t)"- 2 f ( -"- 1) (t)dt 



1 



(«-!)! 



(* - sr ~ l r~ l w -...-(x- a/'® + f ,. 



But 



^ = \l f\t)dt =m-rm 

and therefore 

/(*) «/(© + (« -!)/'(!) + ... 



1 



as required. Observe that, by question 4 above, 



fc-8f , ~ 1 /*-*to+J$, 



i 



/^to) (7 (x-tf^dt = ^r^/ (n) (r?) 



(h-1)! V, 'J* R! 

for some value of tj between x and £. 

• Exercise 13.34 

1 (0 J„ "' J X .,, = f-2(!-x)" 2 ]'- 5 =2-25" 2 ^2as5->0+. 



«/. g 



o V(l -x) 
* x 2 dx 



3\-llX 



(i + x y 



= [-ki+*r 1 io 



111 1 

— — -»■ — as X •* + **, 

3 3(1+ A -3 ) 3 



(iii) Since 

(1 4-x+;c 3 )(l +x 7 ) 
\ +x+x s 

we can find an H such that 



1 as x •* + °° 



Solutions to exercises (13.34) 



221 



1 + x + x 3 < H 



l+x+x s \+x- 



(x>0). 



The existence of the improper integral then follows from proposition 
13.29 (see example 13.30). 

(iv) £"V5f = **-**"*** = .4- 2 ^ +ooas5 - o+ - 

(v)Put 1 -x=y.Then 
r i-8 dx f & — dy _ r 1 dy_ 

y J 6 y 



V§ 



j-i-6 dx i-6 — dy _ r 1 

Jo 1 -y ~ Ji v Js 



1 -x 

and so (v) reduces to (vi). 
(vi)Put.y = 1/z. Then 
i-i dy r 1 zdz _ f 1/5 dz 

J& ~y~ = Jl'6 ~~^~ "Jl Z 

by example 13.33. 



-°°as8->-0 + 



2 / 



= i: 



'•2 + >- 2v — 1 r 1/2 2*- 

dx = 



1/2 -y x(\ -x) 



f 1/2 2x-i r 1 

Ji/2-y v(i -x) J" 



1/2+ y 2x-i 



x(l -x) 



dx. 



Put .v = k - « i" the first integral and jc = \ + u in the second integral. 



Then 



' - -J. ^kTht)"" + Jo «r^r^) d " = °- 

Ho\ 

r 



However, the improper integral does not exist. We have 
-1/2 2sr— 1 f2s> ucto 



'•-^f 



i/2-y x{\ —x) 



(1-Z))(1+U) 



= _r 

Jo 



1 



1 



dv 



l—v 1 + v\ 
o°asy^-\— (exercise 13. 34(1 v)). 



On the other hand 
-1/2+2 2x — 1 



r 

J 1/2 



jc(1 -x) 



dx = +2 



j 



2y urfzi 



(1-0)0 +v) 



- + °° as z ■* t — . 



3 After theorem 13.32 it is enough to establish the existence of the im- 
proper integral 

°dx_ 
x a 



s: 



222 Solutions to exercises (13.34, 14.3) 



But 

r*dx 
x 



rxfa _ 



1 



X 



a-1 X"- 1 



a- 1 



asZ- 



1 -a 
provided thai a > 1 . 

• Exercise 14.3 

1 We have 

log 2 = {* ^>i(2-l)>0. 

Observe that 

log (2") = n log 2 ■* + °° as n -* °° 

and hence the logarithm is unbounded above on (0, °°). Since it is 
strictly increasing, it follows that log* -* + °° as x -*■ + °°. Also 

log (2"") = — n log 2 ■* — °° as « -» °°. 

Thus log a: ■*■— °°asx->0+. 

2 Since the logarithm is concave, 

logj>-logl<(j/- 1) 

for all y >0 (see exercise 12.21(4)). 

Take y =x s . Then, using theorem 14.2(a), 

s log x = log (x s ) <x s — Kx s . 

(i) Given r > 0, put s = \r in the above inequality. Then 

x~ r \o%x = x- r ' 2 (x- s \ogx)<x- rn /s--0asx-+ + c°. 

(ii)Puty=x~ l m(i). 

3 We have F'(x) = log x + x . x~ { — 1 = log x and hence F is a primitive 
for the logarithm. By theorem 13.14, 



| o ]og(\+x)dx =J i log /</? = [t log t-t]t 

= 2 log 2 - 1 = log (4/e). 
As in example 13.17, 



£M ,+e ?H*< 



Hence 



I + x) dx < T - lo 



= i n 



k = i 



k 
n 



Solutions to exercises (14.3) 



223 



> + BH" 



n- \ 

n 



l/n 



^ 



<log {(l+- 



n 



log 



(w + 1 )(« + 2) ... (2« — 1) 



I," 



'<7 



<log 



fa + 1) . . . (2k) 



l/n 



(w(2/i)! 



log., 

2n.n\ 

It follows that 



l/n 



-<tog - <l0| 



'(2«)! 



h! 



l/n 



0<log- 



(2»)! 



4\ 1 
— log — < — log 2 ■* as n -*■ °°. 



4 Consider the function 0: (0, °°) -*B8 defined by 
<p(x) = log* —ax. 

If a < 0, </>(x) -» —oo as x -* + and (p(x) -* + °° as x -* + °° (exercise 
14.3(2)). It then follows from the continuity of that a solution of the 
equation $(x) = exists. 

The case a > is somewhat more interesting. It remains true that 
<p(x) ^ —°° as x -* + but, for this case, 0(.*) ->— °°asjc -* +°° 
(exercise 14.3(2)). Consider the derivative 

<p'(x) = --a. 

From this identity it follows that is strictly increasing on (0, I/a] and 
strictly decreasing on [1/a, °°). Thus 

*(*><* (| -tog~l ° 3) 

witli equality if and only ifx = 1/a. Because <j> is continuous on (0, •*>), 
solutions of <p(x) = exist if and only if the maximum value 0(1 /a) of 
0is non-negative, i.e. 

log(l/a)>l = loge. ( 14 ) 

Since the logarithm is strictly increasing, (14) holds if and only if 
l/a>e,i.e. ea< 1. 

The inequality e \ogx<x(x> 0) is simply (13) with a = 1/e. 

We prove that the sequence (x n ) is decreasing and bounded below by 



224 Solutions to exercises ( 14. 3) 

e. We have 

X„+i = elogx„ <x„ 

and so <x„> decreases. If x n > e, then 

jc„ + i = elogx„>eloge = e 

and so the fact that <x„> is bounded below by e follows by induction. 

Suppose that x n -*■ f as n -*°°. Since x n+1 = e log x n and the 
logarithm is continuous, 

£ = elog£. 

But tliis equation holds if and only if % - e. 

5 Apply theorem 13.32 with/(x) = 1/x. Then 
1.1. .1 r n dt 



A„=l + i + i + ... + i-f^ 

2 3 n Ji { 

and A„ -> y as /? -* °°. 

As in the solution to exercise 6.26(6), 

111 11 

1 + +... + - = — 

234 2n - 1 2n n+\ n + 2 



1 



1 



+ ... + — 
2« 



2 2n 



2 n 



= {log 2n + A 2n }-{log« + A„} 

= log2+A 2n -A n 

■* log 2 + y — 7asn-*°°. 

6 {i)D{\ogx} s =s{\ogx} s - l \lx (ii)0{loglogx} = 

If r 96 1, take x - 1 = -r in (i). Then 



1 1 



j2 r(logx)'" 



dx = 



1 -r 



(log*) 1 



x 

i(log 2) 1 



r-1 



logx X 
(r<l) 



is A" ■* + °°. If r - 1 , we use (ii). Then 
* 1 



I. 



x log* 



dx = [log log xj 2 '"* + °°asx ■* + °°. 



The appropriate conclusions about the given series follow from 
theorem 13.32. 



Solutions to exercises (14.5) 



225 



• Exercise 14.5 

1 (i) Put X = exp x and Y = exp y. By theorem 14.2(f), 

log AT = log X + log Y = x+y. 

Hence 

exp(x+j<) = exp {log XY} = XY = (expx)(exp.y). 

Alternatively, consider 



D 



ex<p(x+y) 



exp x 



exp (x + y).expx — expx.exp (x + v) 



(exp x) 2 



= 



for a fixed value of v. We obtain 
exp (x+y) 



expx 



= c (xGR) 



for some constant c. The choice x = shows that c = expy. 

(ii) Put X = exp x in theorem 14.2(H). 

logA" r = Hog A = rlog (expx) = rx. 

Hence 

exp(rx) = exp (log X r ) = X r = (expx) r . 

2 (i) By exercise 14.3(21), for any s > 0, 
A~ 5 log A^0asA^ + °°. 
Hence 

(log AT r A = {(log A) A~"T r ^ + °° as A ^ + <=•=. 
It only remains to put X = exp x (see exercise 8.20(2)). 
(ii) By exercise 14.3(2ii), for any s > 0, 
A s log A ■* as X -* +. 
(log A)'"A = {(log X)X 1,r } r ■* as A -> +. 
Put X = exp x again and the result follows. 



expx — 1 , expx 
3 (i) lim — = hm —f- = expO = 1. 

x — X x-*0 1 



,.,, .. log(l+*) ,. (1+*)'' _ 1 

(11) hm = hm ; = — — - 

w * -► x * -* 1 1+0 



= 1. 



4 We have 



226 Solutions lo exercises (14.5) 

H\t) = li(t)>H(t). 

Hence 

T* H'(t) c x 

Jo m dt> >* Udt = x (x>oy 

Thus 

[logH(t)] x >x. 

But H(0) = 1 and it follows that 

log//(x)>x 

i.e. h(x) > H(x) > exp x. 

5 We have 

/'(*) = 18.3a; 2 exp (1 8x 3 ) 

f"(x) = 18.6xexp(18x 3 ) + (18.3) 2 x 4 exp(iax 3 ). 

The sign of f"(x) is the same as that of 

4>(x) = 2x + 18.3x 4 = 2x{\ + 27x 3 ). 

Observe that 0(0) = <t>{—\) = 0. Since 

4>'(x) = 2 + 8.27x 3 = 2 + (6x) 3 , 

it follows that <p is strictly increasing for 6x > — 2 1 ' 3 and strictly 
decreasing for 6x < — 2 1 ' 3 . 

Thus is non-negative on the intervals (— °°, — |] and [0, °°) and so 
/is convex on these intervals (theorem 12.19). But is non-positive on 
[—3, 0] and so/is concave on this interval. 

6 Since /increases on [0, °°) 
f(x)<f(k) (k-l<x<k) 
f(k)<J\x) (k<x<k+l). 
Hence 

£, f(x)dx<f(k)< I* 1 max. 

But 

J mdx = £ J ft _ i f(x)dx; J, /(x)dx = £ J ft /(*)<&. 

ft — 1 k — 1 

Apply (he result with/(x) = log x. 



Solutions to exercises (14.5, 14. 7) 

[x\ogx-x\ n < £ \ogk<[x\ogx-x\r 1 

« log n-n< log «!<(« + 1 ) log (rt + l)-(« + 1)+ 1 
nlogn-log«! <«<(«+ l)log(n+ 1) — log/j! 



227 
(15) 



log 



R! 



— <n<log 



(« + ir 



n 



(« + i)" +1 
— < exp /? ^ • 



«! 



n 



(Note that (1 5) involves the use of an improper integral on the left 
hand side. From exercise 14.3(2), x logx -* asx ■* +.) 

• Exercise 14.7 

1 (i)c?* y m exp {(x+y) logo} 

= exp {x log a + y log a} 

= exp (x log a) exp (y log a) (exercise 14.5(H)) 



(ii) (abf = exp {x log {ah)} 

= exp {x log a + x log ft} (theorem 14.2(i)) 
= exp (x log a) exp (x log b) (exercise 14.5(H)) 
= a x b x . 

(m)a~ x = exp {-x log a} 

= (exercise 1 4.5(1 ii)) 

exp (x log a) 

J_ 
~ a x ' 

(iv) Note that, since a x = exp (x log a), it follows that X log a = 
log (a*). Hence 

a xy = exp {xy log a} = exp {y log (a*)} = (a x ) y . 

2 We have 

Z) {exp (x log a)} = log a . exp (x log a) = (logff)a*. 

3 By theorem 8.9 and exercise 14.5(3ii), 



228 Solutions to exercises (14. 7) 
n log(l + xn' 1 ) = x 



JogO+xn- 1 ) 



xn 



■ x as n -* °°. 



Since the exponential function is continuous at every point, it follows 
that 



1 +-) = exp « log 1 +- 



exp x as n 



4 By exercise 14.3(2i), n l log n •* as n -* °°. Hence, 
log n) •* exp = 1 as n -» °°. 



n v " = exp 



5 (i) Consider 
j^~°e' xl/2 dx. 

Take <j)(x) = e~ xn in proposition 13.29. We know that 
^ e~ xn dx = [-2e~* n ] x ->2asX-*°° 
arid 

A similar argument (with tp(x) =e x ' 2 ) establishes the existence of 
r' e-^dx. 

J-*- DO 

It follows that the improper integral 

L— * * 

exists. 

(ii)See §17.3. 
(iii) We have 



Jo ] 



+ AT : 



dx = r£iog(i+x 2 )js 



2\1.Y 



= | log (1 + A" 2 ) ^+°° as *->•+«>. 
Hence the improper integral 
xdx 






+ .v 2 



Solutions to exercises (14.7) 
does not exist, even though 



229 



/. 



-x 1 +*' 



= 0->-0asA'^+°°. 



6 (i) The trick is to differentiate the given equation with respect to x 
keepings constant, and then with respect toy keeping* constant. For 
each x and y, we obtain 

f'(x) = f'(x+y) =/'<» 

and it follows that 

f\x) = c (xGR) 

for some constant c and hence that 

f(x) = ex + d. 

Since /(0) = 2/(0), we must have d = 0. 

(ii) As in (i) we obtain 

/V)/00 - /0)/'0) 

and therefore 
fix) 



= c 



(xGR). 



Hence 

log/(x) = G* + rf. 

Since /(0) =/(0) 2 , d = and hence 

fix) = e cx (x S R). 

(iii) As in (i) we obtain 

/'(*) _f(y) 

y x 

and therefore 

f\x)= C - (*>0). 

Hence 

fix) = clog* +d (x>0). 
Since/(l) = 2/(l),d = 0. 
(iv) As in (i) we obtain 



230 Solu tions to exercises (14.7, 15.6) 

/'(*)/(?) _ f'(y)f(x) 
y x 

and therefore 

/(*) x 

log/(jc) = clogjc + d. 

Since /(l) = f(\f,d = and hence 

f(x)=x c (x>0). 

• Exercise 15.6 

1 (i) The interval of convergence is R because 

|<7 n |"" = — >-0 as«^-°°. 

n 

(ii) The interval of convergence is {0} because 



(n + 1)! 



= (« + 1 ) ** + °° as n 



(iii) The interval of convergence is U. This is most easily seen by com- 
paring the series with the Taylor series expansion for exp x. 

(iv) As (iii). 

(v) The interval of convergence is (—4, 4). The radius of convergence is 
equal to 4 because 

(n+1) 2 (« + l) 



1 

■ — as n -*■ °°. 
4 



(2« + 2)(2#i+l) 2(2« + l) 

Note that the series cannot converge at the endpoints x = 4 and 
x = —4 of the interval of convergence because <<z„4"> is strictly increas- 
ing and hence cannot tend to zero. We have 

» — = r > 1 (w = 1 1 2, . . .). 

a„4" (« + J) 

(vi) The interval of convergence is [— 1, 1). The radius of convergence is 
equal to 1 because 

v« 



V(» + 1) 



■* ] as n -*■ °°. 



The series does not converge for x = 1 . 

The series converges for* = — 1 by theorem 6.13. 



Solutions to exercises (15.6) 



231 



2 To employ exercise 13.26(6), we need the nth derivative of the ftlBO 
tion/:(-l,°°)-»-R defined by /(*) = log (1 + x). We have 



fix) = 



1 +x 



f\x) = - 



(i+*r 



f<"\x) = (n-l)\ 



(-1) 



n-1 



(1+x)"- 



(A formal proof would require an induction argument.) These deriva- 
tives are evaluated at x = and from exercise 13.26(6) 



g(i+^) = /(o) + f [ /'(0) + ... + ^ i / ( "- 1) (o)4-^ 



lo 



where 



= x + - — ■ • • + 



(-1)" 



(H-l) 



+ En 



E » - ==niir *-« , -w« 



(8-1) 

1 



dt 



— 2— re 

(n-l)! Jo v 



x-ff^H-l)! 



(-1)"" 1 



dr. 



(1 + 0" 

These formulae hold for any x > - 1 . We are concerned with the values 
of x for which E n ->• as « -> <*>. We therefore have to consider 



I4J = 


Jo (1+0 




ffO<x<l. 


^=i; < ( 1+ 'V < " < i'» ( '-' r '* 


x" „ 

= ► as m -* °°. 

n 


If-Kx<0, 


f o (r -*)""' 


It is ea: 


ily seen that 





232 Solutions to exercises (15.6 j 
.t—x 



0<- 



-<-x (-Kx<t<0) 



1 +t 
and it follows that, for - 1 < x < 0, 

\E n \A x (-xf * — 

= (— ^)" _1 [— log (1 + *)]-»■ as w ->• °°. 
The series expansion for log 2 is now obtained simply by setting.* = 1. 

3 To obtain the form of the remainder as given in theorem 1 1 .10, we 
have to write x = 7? in the formula for/ {n) (x) in the previous question. 
This yields 



E„ = 



(-D 



n-l 



1+TJ 



where r\ lies between and x. If < x < 1 , there is no problem. Since 

1 + 7J>1, 

n 
If — 1 < j: < 0, we can assert no more than 

But 

—x 

1+x 



R \1 + X 



>1 (-1<jc<-|) 



and hence we are only able to show that the remainder term tends to 
zero for values of x satisfying — ^ < x < 1 . 



4 We have 



Rn = 



1 



1 



(n + 1)! (« + 2)! 



1 



1 + 



1 



■ + • 



1 



< 



(n + 1)! 1 (n + 2) (n + 2)(« + 3) 
1 



+ . . . 



(« + 1)! 

1 
(« + 1)! 



(n +2) (n + 2) 2 
1 



+ .. . 



!-(« + 2) 



i-i 



Solutions to exercises (15.6) 
1 n + 2 



25.? 



(« + !)! n + 1" 

From the power series expansion for e x with X = 1 , we know that 



1 1 1 

0<e-l+— + — + ...+ — 



1! 2! 



H! 



= R„. 



We have to take n = 4. Observe that 
16 16 1 



R t < 



5! 5 



120 5 



100 



5 Suppose that e = m/n where m and n are natural numbers. Then 



1! 2! 



Rj 



is a natural number. But, by the previous question, 
n! (n + 2) 



0<7 = n\R„< 



(n + l)!(n + 1) 
( W + 2 ).<1 



(n + l) 2 



provided that n is a natural number. But there are no natural numbers 
in the interval (0, §) and hence we have a contradiction. (Note that 



n + 2 



1 



■ + ■ 



1 



(n + l) 2 (r + l) 2 ' (n + 1) 
and hence is obviously decreasing.) 

6 We prove by induction that, for any x 4* 0, 

f«Xx-) = P n ^\e-^ 



(16) 



where P„ is a polynomial of degree 3n. For n = 0, there is nothing to 
prove. Given that (16) holds, 



/ (n + °(*) = K 



-^K^ + Mt)-^ 



= ^HH& : 



Observe that the polynomial P n + 1 defined by P n+1 Q>) = -y 2 P' n (y) + 
2y 3 P n (y) has degree 3n + 3 provided that P„ has degree 3n. This com- 
pletes the induction argument. 



234 Solutions to exercises (15.6, 15.10) 



We next consider the existence of the derivatives / <n> (0). Again we 
proceed by induction. We show that/ <n) (0) = (n = 0, 1, 2, . . .). This 
is given for n = 0. On the assumption that/ ( " _1> (0) exists and is zero, 

/<"-»(*)-/t"-'>(0) i M _ v< 

x x ""' \x 



It follows that 



lim 

x -» 



lim j>/ViO)e~ 5 ' = 



by exercise 14.5(2) (exponentials drown powers). The same argument 
also applies with the left hand limit. It follows that/ (n) (0) exists and is 
zero as required. 

Since /„(0) = (n = 0, 1, 2, . . .), the Taylor series expansion of/ 
about the point 0, is 

+ + + + + .. . 

This converges tof(x) only when x - 0. 

• Exercise 15.10 

1 The radius of convergence R of the power series 

fix) = I «("-l)--(«-"+l) JC „ 



n=0 

is given by 



n! 



rt! 



^ „™. (« + l)! |a(«-l)...(a-/i + l)| 

.. |a-n| 

= lim = 1. 

«-»■« « + 1 

For |jc| < 1 we may therefore apply proposition 15.8 and obtain 

= «a-l) »-» + !) 

Hence, 

(1 +*)/'(*) 

- a(tt-l)...(a-w+l) ~ tt(a-l)...(tt-H+l) „ 

A (»-D! 4l («-D! 

" I a(a— 1) ... (a— w) a(a — 1) . .. (a — n+ 1) 

= a + ) ( 1 



Solutions to exercises (15.10) 

a(a — 1) . . . (a — n) 



235 



= «+ I 



(«-!)! 



U-L 



n a—n 



= a 



i+ - tt(tt-l)...(a-/i +1) ^,1 



«! 



It follows that, for \x\ < 1, 

D{(\ +xy a f(x)} = -a(I +x)- a -'/(x) + (l +x)- Q /'W = 

and therefore 

f(x) = c(\+xf (|x|<l) 

for some constant c. But /(0) = 1 and hence c = 1 . 

2 The radius of convergence R is given by 

1 r (n + l) 2 , 
— = hm ; — = 1. 

R n - - n 2 

We know that 



n = 

Hence, usi 

oo 



1 



— X 



(W<1). 



Hence, using proposition 15.8, 
1 
(1 -^) 2 



CM<1) 



2 



X «(«-l)x"- 2 = — — - 3 (|x|<l). 

Therefore, 

£ «V = J {n(n-l) + n}x n 

n=l 
= X 2 . £ «(«-l)x"- 2 + X. J BX"" 1 

2x 2 



n = 



n = 2 
.2 



+ 



(1-x) 3 (1-x) 2 

2x 2 +x(l + s) _ x(l+x) 
(1-x) 3 = (1-x) 



3 (W<1). 



3 The power series 



236 Solutions to exercises (15.10) 



_y + i 



has the same radius of convergence as the given power series because 



lim sup 



lim Lap \a n \ 



(« + I) 

By proposition 15.8, if x e / but is not an endpoint, 



F'(x) = f 



n% (« + 1) 



.(n+l)x n = f{x). 



Hence 

fo-f(o) -£/&#*. 

Observe that F(0) = 0. 

4 The function/: (— °°, l)->-|R defined by 
log(l-*) 



fix) = -- 

X 

has the Taylor series expansion 

/(*) = 1+I + J + J+--- (-1<*<1). 

(Note that formula (1 7) does not make sense when x = 0, but it is 
natural to define /(0) = 1.) 

If < y < 1 we obtain from the previous question 

\ f(x)dx=y+f 2 +f 3 +... 

and hence 



(17) 



r 1 s, v" 

lim /{*)<& = lim £ V 



But the final power series converges for .y = 1 , and from proposition 
15.8 we know that the sum of a power series is continuous on its inter- 
val of convergence. Thus 



-/. 



' log(l-*) 



1 



x „=i n 



5 We consider the power series 



*(*)= I («»-*„)(* -8". 

n=0 



Solutions to exercises (15.10, 16.3) 

Ifgix) = (x € /), then, using proposition 15.8, 
= g(£) = a -b 
= g'm = l(«i-*i) 
= /(£) = 2(a 2 -b 2 ) 
= *'"«) = 3.2(a 3 -6 3 ) 



257 



and the result follows. 
6 By proposition 15.8, 

n = 

From the previous question it follows that 
a n _, = «a n («= 1,2, .. .) 
as required. But then 

1 1 

22 = ^1 = — «0 

and, in general, 

a „ = ^ (« = 1,2,...). 



n\ 



Thus 



/(*) = "o I -T = «o**. 



n=0"l 

• Exercise 16.3 

2 4 

1 (i)cos0= 1--+--...-1 

3 5 
(ii)sinO = 0- — + — -. ..= 

(,-xf (~xf 

(iii)cos(-x) = i--^r + -^r 



— . . . = cos X 



238 Solutions to exercises (16.3) 



(iv)sin(-*) = (-*)- 



(-*) 3 , (-*) 5 



3! 



5! 



. . . = — sin x. 



We appeal to proposition 15.8. Then 



Dcosx = D 1 + 

2! 4! 



_ 2* 4* 3 6x s 
2! 4! 6! 



Dsin* = D{x — — + — — , 



= -x+ 

3! 5! 



. . . = — sin x. 



3! 5! 



= '"i + Ij----" 5608 *- 



2 We have 

£'(*) = cos (x+y) — cos x cos y + sin * sin y = ft (*) 

«'(*) = — sin (x +7) + sin* cos j> 4- cos* sin 7 = —g(x). 

Hence 

D{g(xf + h{xf) = 2g(x)g\x) + 2h{x)h\x) 

= 7g(x)h(x)-2h(x)g(x) = 

for all x. It follows that 

g(xf + h(xf = g(0) 2 + h(0) 2 = 

for all x. Since {#(*)} 2 > and (M*)} 2 > 0, we may conclude that, for 
all values of x, 

m = 

and 

«(*) = 0. 

3 (i) 1 = cos(*-*) = cos*.cos(-*)-sin*. sin(-x). 

(iii) < cos 2 * < cos 2 * + sin 2 * = 1 . Hence |cos x\ < 1 . Similarly 
|sin x| < 1. 

(iii) sin 2x = sin x. cos x + cos x. sin x 

(iv)cos2* = cos x. cos* —sin*. sin*. 



. ... .. sin* cos* 

4 (1) hm = lira = cosO = 1. 

x -* * X -► 1 



Solutions to exercises (16.3, 16.5) 



239 



1 — cos * sin * 1 

(ii) lim ; = lim — - = — . 

x-*o x *-»o 2* 2 

5 By the mean value theorem, there exists £ between and * such that 
sin * — sin 



*-0 



= cos £ 



and we know, from question 3 above, that |cos £| < 1. 

The convergence of E~ =1 sin (l/« 2 ) may be deduced from the com- 
parison test because 



sin 



\n 



<4: («=1,2,...). 



Alternatively, we may appeal to exercise 6.26(2). From Question 4(i) 
above 



lim 



sin(l/« 2 ) 



= l; 



sin(l/«) 
lim : = 1. 



I//1 2 "*-*"" 1/n 

The latter result shows that S~ = , sin (1/n) diverges. 

6 From exercise 13.26(5) it follows that, for some % between m and n, 
r" sin r 1 rJ 1 r " 



r" sinr 1 r( , , 1 f" . 

dt = — sin t dt + — sin t 

Jm [ m Jm n Jf 



dt 



= — (cos m — cos ?} + — {cos if — cos n\. 
m n 



Hence 

r" sin t 
Jm ( 



dt 



< 



m 



Let e > be given and choose N = 4/e. Then for any n>m>N, 
r" sin / 



f" sin / r m sin t 

, dt-\ 

Ji t Ji f 



dt 



4 4 



and it follows that we are dealing with a Cauchy sequence. Hence 

f sin t 

lim 

„-»°o Ji ? 

exists (theorem 5.19). 

• Exercise 16.5 

1 From the discussion of § 1 6.4 we know that cos * > (— \ti < * < \n ). 
Since sin (* + jn) = cos*, it follows immediately that 



240 Solutions to exercises (16.5) 

sinx>0(0<x<7r). But 
D(cosx) = — sin* 

and thus the cosine function decreases on [0, »r]. Because cos (x + ir) = 
— cos x, we may also deduce that the cosine function increases on 
[ir, 2ir] . Further, 

D 2 (cosx) = — cosx. 

Thus Z> 2 (cos x) < (— %n < x < \it) and so the cosine function is con- 
cave on [ _ |ff,|jr]. Because cos (x + n) = — cos x, D 2 (cos x) > 
(jn < x < §7r) and so the cosine function is convex on [\it, §jt] . 



2 We have 



tan (x + n) = 



sin (x + it) 



sin x 



cos (x + n) — cos x 
(provided that cos x ^ 0). Also 

cos x. cos x — sin x (— sin x) 



= tanx 



.Dtanx = 



cos 2 x 



cos 2 * 



(provided that cos x ¥= 0). Since cos x does not vanish on (— \n, |jr), it 
follows that tan x is strictly increasing on (— § jt, ^it). We have cos x -► 
as x •* \it — and sin x •* 1 as x -> ^w — . Since the sine and cosine func- 
tions are positive on (0, |ff), it follows that 

sin x 

tanx = »-+ 00 asx-+57r— . 

cosx 

Similarly for tan x -*■ — °° as* ■* — ^rr +. 

3 We have 

£>sinx = cosx>0 (— |»<x<Jff) 

and hence the sine function is strictly increasing on [ — 5^, \n\. Since 
sin (— \t() = — sin (£*•) and sin (\it) = 1 , it follows that the image of 



.v = arcsin y 





Solutions to exericses (16.5) 
1 1 



241 



D (arcsin v) = , 

D sm x cos x 

where x = arcsin .y, i.e. sin x = y. Since cos 2 x + sin 2 x = 1 , we have 
cosx = ±V(1 -sin 2 x) = ±V(1 -y 2 )- 

We take the positive sign because we know that the arcsine function 
increases. Thus 

D (arcsin y) = 



1 



(-Ky<\). 



V(i -y 2 ) 

We have 

Dcosx = -sinx<0 (0<x<tt). 

Also cos (0) = 1 and cos (ir) = — l. Hence the image of[0, n] under the 
cosine function is [— 1 , 1 ] . 





£>(arccosj>) = 



I 



1 



-1 



Dcosx — sinx \/(l ~y 2 ) 



We take the negative sign because the arccosine function decreases. 
Note that 



cos 



(x + ^tt) = — sinx = sin(— x). 



If x lies in the range [— \-n, \ti] we may write — x = arcsin^. But 
then 

cos (x + ^7r) = ;'. 

Since x +\n lies in the range [0, ir) it follows that 

x + %ir = arccos>>. 

i.e. \u = arccos>> + arcsin y. 

4 It follows from question 2 above that the tangent function is strictly 



242 Solutions to exercises (16.5) 

increasing on (— \n, \-n) and tan x -*• + °° as x •* \n — and tan * 
as x-*— \tj+. Thus the image of (— \ti, \ti) is R. 




3' = lan .v 



jrX = arctan v 


-w 


";■ 



Z) arctan J» = 



1 



= cos 2 * (y = tan *). 



D tan x 

But cos 2 x + sin 2 * = 1 and so 1 + tan 2 * = 1/cos 2 *. Thus 
1 1 



D arctan y = 



1+tan 2 * " 1 + y 



2 ' 



Finally, 
C x dx 



r- 



+ * 2 



= [arctan *] = arctan X-*- — as X-* + °°. 



5 Differentiate with respect to* keeping y constant and then with res- 
pect toy keeping* constant. If*y < 1, 

f {x) m f ( x+ y\ f ] (i -*y)-(x+y)(-y) 



(\-xyf 



Hence 



f'(x) - /' 



Also 



, (x+y 

1 —xy 



1 +y 2 



x +y 



(1 -xy)- 



1 +* 2 



,2 ' 



'<» = '" \i -*,/ id -xy?\ ■ 

Thus /'(*)(! +* 2 ) =f'(y){\ +y 2 ) (xy < 1). Hence 



fix) = 



1+* 2 



(Xi 



Therefore /(*) = c arctan * + d (* E M). But 2/(0) = /(0) and so 
d = 0. 



Solutions to exercises (16.5) 243 

If* = y, we obtain 

»-£'(&) (J<>1) ' 

We apply this formula with/(*) = arctan *. Since arctan y -* j7r as 

y-> + °°> 

( 1 ( 1* 

arctan 1 = lim {arctan*} = hm —arctan I- _ j 





= U- 


The 


power series 


* — 


x 3 x 5 

— + . . . 

5 5 



is easily seen to have interval of convergence (— 1 , 1 ] (for * = 1 , use 
theorem 6.13). We apply proposition 15.8. Then, for |*| < 1, 



D 


* 3 X 5 

arctan *—* + — — — +... 




i | v 2 „4 | 


- ]+x2 11**1 ... 




i i 


1+* 2 1+* 2 


It follows that 


* 3 * s 
arctan * = C + x— — + — — . 



(W<1). 



But arctan = and so C = 0. 

We know from proposition 15.8 that the sum of a power series is 
continuous on its interval of convergence. Hence 

\-n = arctan 1 = lim_ {arctan*} 





x 3 x s 


= lim 


*- T + --... 


* -» t - 


3 5 



— I 3 + 5 

6 (i) Consider the sequence <*„> defined by 
* n = {i7r + 2«7r}-'. 
Then *„ -*■ as n -* °°. But 
/(*„) = sin(|ir + 2n7r) = 1 -/►/(()) as «• 



244 Solutions to exercises (16.5, 1 7.4) 




(ii) We have \g(x)\ < |x| and hence g(x) -* as x -> by the sandwich 
theorem. However, 

g(0+x n )-g(0) 
*•> 

g(0-x n )-g(Q) 



•* 1 as n •*• °° 



—x„ 



-*■ — 1 as n •* °°. 



(iii) We have \h(x)\ < x 2 . Hence 
h(x)-h(p) 



-0 



< Ixj ->0 as;t->-fJ 



and hence m'(0) = by the sandwich theorem. 

• Exercise 17.4 

1 We have < sin x < 1 (0 < x < ±tt). Hence 

0<sin n + 1 x<sin"x 

and therefore 

0</ n+1 </ n (n = 0,1,2,...). 

To obtain the recurrence relation nl n = (h — l)/„- 2 , we integrate by 
parts. For n > 2, 

/„ = J o sin"* 

= sin x. sin" 'jtdx 

= [-cosx.sin"- 1 x]S /2 +n cosx.(n-\)sin n - 2 x.cosxdx 



Solutions to exercises (1 7.4) 



245 



- (n - 1) J"' (1 - sin 2 *) sin"- 2 * <£c 

= (n-1 ){/„-»-/„}. 
i.e. n/„ = («-l)/„-a (« = 2,3,...). 
It follows that 

hn Jin-i 2m + 1 



l<-^< 



■-*■ 1 as n -> °°. 



An + l ^2n + l 2« 

2 Use (18) of the previous question. We obtain 

r (2m ~ 1) . (2h-1)(2w-3) t 

/*„- 2/? li»-a- 2 „ (2w _ 2 ) i2 '- 4 

_ (2«-l)(2n-3)...l 
2k(2m-2)...2 ° 

_ 2m(2m - l)(2w - 2)(2« - 3) ... 2.1 r*« 
(2m) 2 (2h-2) 2 ...2 2 '• lo 

-J^f c»-o,,. 2> ...,. 



2m 2«(2n-2) 



2« + l '"' (2m+1)(2m-1) 

2h(2w-2)...2 

(2m + 1)(2m - 1) ... 3 ' 

(2m) 2 (2m - 2) 2 . . . 2 2 - »« 



/ 2 n-3 



•f 
Jo 



(2m + 1)(2m)(2k-1)...3.2 
(2 B «0 2 r ..,„/. (2 n «!) 2 



(18) 



[-cosx]5 /2 = 



(2m + 1)! 



sin x <& 



(k=0,1,2,...). 



(2n + 1)! 
It follows that 
l 2n _ (2k)! n (2n+l)\ -n {(2m)!} 2 



/a„ + i (2"«!) 2 " 2 " (2"m!) 2 2 v ~" * y {2"m!}' 

in i n {co) 2n+i/ y 2 "} 2 

~2 ( J {2 n Cn" +1/ V n } 4 

,2 4 "2.M 4 " + 1 e- 4 " 



7T 



= ^(2"+OC" a " 2 4n"' B 4n^ e ^n 



2m+ 1 



n 



C 2 -*-2nC~ i asn^-°°. 



By the previous question, 27rC 2 = 1, i.e. C = y/(2ir). 



246 Solutions to exercises (1 7.4) 



3 The fact that (d n — (12m) '/increases was demonstrated in §17.2. As in 
§17.2, 

x 1 x 4 x 6 I 1 

*-*♦,-/» -j+ 7 + 7 +... X- — r 



Hence 



1 



x* 1 
d n -d n + l >- = 3 (2w + 1)2 

But 



1 



12m 2 + 12m + 3 



1 



1 



12m + 13- 12m -1 



12m + 1 12(/i + 1)+1 (12m + 1)(12« + 13) 

_ 12 

" (12m + 1)(12m+ 13) ' 



It follows that 

i 



d n ~ 



1 



12m + 1 



- *,♦»- 



l 



12(« + 1)4-1 



> 



(12m + 1)(12m+ 13) - 12(12m 2 + 12m + 3) 



3(12m+1)(12m+13)(2m + 1) 2 
The numerator is equal to 

12m(14-12) + (13-36) = 24«-23>0 (m = 1,2, ...) 
and therefore the sequence (d„ - (12m + l)""') decreases. 



We know that d n ■+ d as n -> °° and therefore 

(m = 1,2,...). 



d n ~ -T < d < d n 

12m " 12m + 1 



Taking exponentials and noting that C = exp d = s/(2ir), we obtain 



'<- 



^g-l/Oln + l) 



n< 



as required. 

We may deduce that 



1(1 _„-l/(12n + D 



m!(1 



)< m! - y/(27T)n n + U2 e-" < m!(1 -e" 1 ' 12 "). 



Since e x — 1 is approximately equal to x for small values of x (Why?), 
the maximum possible error in approximating to n! is about 



12m 



n\. 



Solutions to exercises (1 7.4) 



247 



When m = 100, this is small compared with m! (though, of course, it will 
be a very large number compared with those usually encountered). 



4 Consider, for < 5 < A, the value of 
j s f->e-<dt = 



— e 
x 



-t 



+ 



r. 



- e' 1 dt 



= -{A x e- A -8 x e- s } + - C fe'* dt. 
x x J & 

Observe that A x e' A -* as A -* +°° (exponentials drown powers) and 



5 x e~ s -* as 5 -* +. It follows that 



rw = -r(x + l) 

as required. 



5 Let0<a<a<jc<^<6</3andlet0<5< A. Then 



{!>->'**< 



f y-i „-t 



e' c dt 



<J" S I**" 1 -/*"*!*"**. 



We may apply the mean value theorem and obtain 
r*-i_,y-i 



x-y 



= Qogtytt-* 



(19) 



(20) 



for some £ between x andy. 

For any r > 0, t~ r log f -*■ as t ■* + °° and r r log r ■* as r ■* +. It 
follows from (20) that we can find an H such that 

\t x - 1 -t y - 1 \<H{t°- 1 + t p - x }\x-y\ 

and hence it follows from (19) that 

TO - r(y)i < h\x -j>i{r(oo + tm) 

and the continuity of the gamma function then follows from the sand- 
wich theorem. 

6 We show that log V{\x + \y) < \ log P(x) + \ log T(y) and appeal to 
exercise 12.21(6). By the Schwarz inequality (theorem 13.25), if 
0<5<A, 

j" 5 A ,U* y -2),2 e - t J m | 6 A (^-0/2 e - ( /2 )( ^-l)/2 e -(/2 )A ) 



< 



t*-\ „-t 



e~' dt 



St"-"- 



dt 



248 Solutions to exercises (1 7.4, 1 7. 7) 



Thus 



and the result follows. 

• Exercise 17.7 

1 Make the change of variable — t = log u in the integral 

.A 



j f^e-'dt. 



2 We use the inequality 

+ nfn\ = Cv + n) y r(n + 1) < TO + « + 1) < (" + 1)T(« + 1) 

■ (n + l) y w! (21) 

established in the proof of theorem 17.6 for < y < 1 and « € M. 

Given x, we let n + 1 be the largest natural number satisfying 
n + 1 < x and write x = y + n + 1 . Then < y < 1 . 

Consider 



r(*) 



< 



(« + l) y «! 



(« + l)V(2w)n W" 



y/(2Tt)x x x- U2 e- x y/(2ir)x"x "' 



X x- U2 e- X y/(2n)x x x- in e- 

(/! + l)WVV" 



(y+n+\) y+ "(y+n+iy ,2 e 



l/2„-y-n-l 



■* 1.1. 



jy + « + 1 
1 



n 



1 + 



y + i 



,y + i 



>> + « + lj 

e y * ' = 1 as n -*■ °°. 



,y + i 



A similar argument for the left hand side of inequality (21) com- 
pletes the proof. 



3 We have 

(log(l +z)-z 



lim 

*-* o 



Consider 



= lim 

z -► 



= lim 

Z-+0 



d+z)-'-i] 



2z 
-(1+z) 



■i) 



^ 



Solutions to exercises (1 7. 7) 

y/u f u (U + Xy/u) - ^ (« + Xy/uf^ <T" "* V " 
y/u{u+Xy/uy- 1 e- u -*^ u 

V(27r)« u w- 1/2 e- u 
\u-l 



249 



V(2tt) " I y/u 



1 



-l 



»i 1+ ^j ex p(" lo 8( 1+ ^l 



Write z = x/y/u. Then 

U log (l + ~J -xy/u = ^ log (1 + Z) - j 

log(l + *)-* 



-»■ — £x 2 as z -* 



and the result follows. 



4 We use proposition 13.29 to check that the improper integral exists for 
x > and y > 0. The inequalities 

t x -\i-ty- 1 <t x - 1 (0<r<i) 
t x -\\-ty- 1 <(i-ty- 1 (o<r<i) 

suffice for this purpose. The fact that B{x, y) is a continuous function 
of x (and of>») is proved in the same manner as exercise 17.4(5). The 
fact that its logarithm is convex is proved in the same manner as exer- 
cise 17.4(6). 

5 After the previous question, we need only show that /(l) = 1 and 
f(x + l) = xf(x). Now 

m = T ^B(l,y ) = yj;\l-t) y -' d t = l. 



ro) 

Also,ifO<a<0<l 

-0 



-r*-(l-f) 3 

y 



+ f* xf- x -{\-tf dt. 

a Ja y 



j a m-ty-'dt = 

It follows that B(x + l,y) = xy~ l B(x,y + 1). But 

B(x,y + l) = r*V- , (l--/f<& = r^-'il-ty-^l-Odt 

J — *■ J - * 

= fi(*^)-5(x +!,>-). 



250 Solutions to exercises (1 7. 7) 
Hence 



B(x+l,y) = -{B(x,y)-B(x+l,y)} 



1 +- B(x + \,y) = -d(x,y) 

y y 



B(x+l,y) = 



x + y 



B(x,y). 



It follows that 



*+»-&£g&** + U* 



= (* +y) 



r(y) 

r(x +y) x 



B(x,y) = xftx). 



T(y) x+y 

Since /satisfies all the conditions of theorem 17.6, it follows that 
f(x) = r(x) ix > 0) and therefore 

Vt " Tix+y) 
6 Take x = y = § in the previous question. Now 

O/l jU - r & r & _ p/K2 

»Q,0- r(1) -I ( 2 ). 

On the other hand 



B &b-Loy/ua-t)\-L*o 



dt 

Thus r(|) = \A- 
Write t = x 2 ll. Then 



1/2 2 sin cos 6 d6 



sin flV(l - sin 2 0) 



= ff. 



Jo 
Hence 



V2 dx = "(a/r^Vd* = --r/*» = 



Jo ( 2? )" 



V2 r ® = V2 



£ 



= 2../- - V(2ff). 



SUGGESTED FURTHER READING 



/. Further analysis 
J. C. Burkill and H. Burkill, ^l Second Course in Mathematical Analysis 
(Cambridge University Press, 1970). 

This book is notable in the clarity of its exposition but is perhaps a little old- 
fashioned in its treatment of some topics. 

W. Rudin, Principles of Mathematical Analysis (McGraw-Hill, 1964). 
This is a well-tried and popular text. The treatment is rather condensed. 

T. M. Aposto], Mathematical Analysis (Addison-Wesley, 1957). 
This is a useful text, especially in its treatment of vector methods. 

R. V. Churchill, Complex Variables and Applications (McGraw-Hill, 1960). 
All the books above include a discussion of complex analysis (i.e. analysis 
involving the entity i = V — 0. b ut this book is a particularly easy introduction 
to the topic. 

H. L. Royden, Real Analysis (Macmillan, 1968). 

This is an excellent book. The choice of material and manner of presentation has 
been carefully considered. Its treatment of the Lebesque integral is particularly 
good. It is, however, a book which makes considerable demands on the reader. 

G. F. Simmons, Introduction to Topology and Modern Analysis (McGraw-Hill, 

1963). 

The title of the book describes its content. It is very well-written indeed. 

2. Logic and set theory 
E. A. Maxwell, Fallacies in Mathematics (Cambridge University Press, 1963). 
This is for entertainment purposes. But it also contains some very instructive 
examples. 

N. Ya Vilenkin, Stories about Sets (Academic Press, 1968). 

This is another entertaining book in which the paradoxes of the infinite are 

explored at length. It is highly recommended. 

P. R. Halmos, Naive Set Theory (Van Nostrand, 1960). 
This is the standard reference for elementary set theory. 



251 



252 Suggested further reading 

K. Kuratowski, Introduction to Set Theory and Topology (Pergamon Press, 

1972). 

This is a very good book which covers some difficult material. 

A. Margaris, First Order Mathematical Logic (Blaisdell, 1967). 

Formal mathematical logic is a fascinating and expanding field. This book is a 

good introduction. 

3. Foundations of analysis 
E. Moise, Elementary Geometry from an Advanced Standpoint (Addison-Wesley, 
1963). 

The real number system was developed largely in response to geometric needs. 
This fascinating and clearly written book describes this process and many other 
topics. 

E. Landau, Foundations of Analysis (Chelsea, 1951). 

This remains an excellent and simple account of the algebraic foundations of the 

real number system. 









NOTATION 



e.* 


1 


f(x) -* I as x ■* | 


- 


74 





1 


/(x) ■* / as x -» | 




75 


c 


2 


/(x) -»4-°°asx- > -£ + 


82 


N.Z.Q.R 


2 


/(x) -» / as = ■* — 


oo 


82 


oo 


3 


fai WB 




91 


>,>,<,< 


3 


0(h) 




92 


M 


10 


/"(?). D 2 m) 




92 


d(x,y) 


11 


8y, 5x 




93 


sup, inf 


15,70 


± 




93 


max, min 


15 


dx 






(fi,b),(a,b),[a.b) 




df 




93 


[a,b], (a, °°), [a, °°) 




dy, dx 




94 


(-«►,&),(-«",*] 


1 


/«+),/(?-) 




108 


d(l S) 


19 


S(P) 




no- 


x„ -► / as w -*• °° 


28 


f"f(x)dx 






lim x n 

n -» oo 


28 




no 


x n "* + °° as n "* °° 


38 


| f(x)dx,\ 

JO J-*a 


f(x)dx 


132 


x n •* — °° as n •* °° 


39 






lim sup x n 


48 


logx, Inx 




137 


n-»M 




e 


137 


,141 


liminfx n 


48 


expx 




140 


oa 




a 1 ', e x 




141 


E« n 


53 


oo 






n = l 




I a n (*-£)" 




143 


f.A*B 


64 


n=0 






fix) 


64 


sin x, cos x 




151 


f(S) 


65 


7T 




153 


f + g.W.fg.f/g 


68 


tanx 




154 


f°g 


68 


arcsin x, arccos x, arctan x 


154 


r 


68 


a n ~ b n 




156 


f(x) ->lasx->Z + 


74 


T(x) 




157 



253 






INDEX 



Abel's lemma, 170 
Abel's theorem, 16 
absolute convergence 

of series, 60 

of power series, 144 
absolute value, 10 
analytic function, 145 
anti-derivative, 124 
arccosine function, 154 
Archimedian property, 20 
arcsine function, 154 
arctangent function, 154 
area, 13,119 
arithmetic mean, 8 

backwards induction, 23, 25 
beta function, 160 
binomial theorem, 25, 106, 149 
bounded 

above, 13 

below, 13 

function, 70 

set, 13 
bounds, 13 
Brouwer's fixed point theorem, 89 

Cauchy mean value theorem, 104 
Cauchy-Schwarz inequality, 7, 13 
Cauchy sequence, 50 
closed interval, 17, 19 
cluster point, 52 
combination theorem 

for functions, 79 

for sequences, 30, 161 
compact interval, 17, 52 
comparison test, 58 
composite function, 68 
composition, 68 
concave functions, 1 14 
conditional convergence, 60 
continuity, 84 et seq. 



at a point, 77 

on an interval, 84 
continuity property, 86 
continuous, 77 

on left, 77 

on right, 77 
continuum property, 141 et seq. 
convergence 

of functions, 74 

of sequences, 27 

of series, 53 
convex functions, 1 14 
cosine function, 151 

decreasing 

function, 108 

sequence, 34 
degree 

of polynomial, 25, 67 
derivative, 91, 95 
differentiable, 91 
differential, 93, 95 
differentiation, 91 et seq. 
distance 

between points, 1 1 

between point and set, 19 
divergent 

functions, 82 

sequences, 38 

series, 54 
domain of function, 64 

element of set, 1 

empty set, 1 

Euler— Maclaurin formula, 134 

Euler's constant, 1 39 

exponential function, 140 

function, 64 et seq. 

gamma function, 157 



255 



256 



Index 



geometric mean, 23 
gradient , 9 1 
graph, 64 

harmonic mean, 8 
L'Hopital's rule, 104 

image 

of point, 64 

of set, 65 
improper integral, 132 
increasing 

function, 108 

sequence, 33 
induction, 22 
inequalities, 3 
infimum, 15,71 
infinitesimal, 94 
integer, 2 
integral, 120 
integral test, 135 
integration, 1 19 el seq. 

by parts, 129 
intermediate value theorem, 87 
interval, 16 

of convergence, 144 
inverse function, 68, 1 10 
irrational number, 2, 9 

Leibniz's rule, 97 
limit of function, 74 

of sequence, 27; inferior, 48; 
superior, 48 
limit point of set, 52 
local maximum, 100 
local minimum, 100 
logarithm function, 137 
lower bound, 14 

maximum, 15, 71 

mean value theorem, 102 

minimum, 15, 71 

Minkowski's inequality, 8,10 

modulus, 10 

monotone 

function, 108 

sequence, 34 

natural logarithm, 138 
natural number, 2, 20 
Newton-Raphson process, 117 
nth root test, 59, 144, 164 



open interval, 17, 19 
oscillating sequence, 39 

partial sum, 53 

partition, 120 

periodicity, 153 

point of accumulation, 52 

polynomial, 25, 66 

powers, 141 

power series, 143 et seq. 

primitive, 124 

principle of induction, 22 

quadratic equation, 6 
radius of convergence, 144 
range 

of function, 65 

of sequence, 27 
rational function, 67 
rational number, 2 
ratio test, 59, 144, 164 
real number, 2 
Riemann integral, 128 
Rolle's theorem, 102 
roots, 6, 112 

sandwich theorem 

for functions, 80 

for sequences, 3 1 
sequence, 27 
series, 53 
set, 1 

sine function, 151 
slope, 91 

stationary point, 101 
Stirling's formula, 156, 160 
strictly increasing or decreasing 

function, 108 

sequence, 34 
subset, 1 
sum of series, 53 
supremum, 15, 70 

tail of series, 58 
tangent function, 154 
Taylor polynomial, 97 
Taylor series, 145 
Taylor's theorem, 105 
tends to a limit, 27, 74 
term of sequence, 27 
transcendental number, 10 



1 



Index 



triangle inequality, 10 
trigonometric function, 1 50 

upper bound, 1 3 



257 



uniform convergence, 147 
unbounded 

set, 14 

function, 71 



Jacket design by Ken Vail 
Printed in Great Britain 



Surfaces 

H.B.GRIFFITHS 

Professor Griffiths relates and explains toplogy in terms of everyday experience. 
Beginning with the question, 'What do we mean by surfaces?' he goes on to 
build up a set of definitions based on an examination of three-dimensional 
models and drawings, and only then to relate these to the more advanced formal 
mathematics. He thus provides the background for readers to inject spatial 
concepts into their mathematical thinking. The text is well illustrated with toned 
drawing which create a three-dimensional effect where appropriate and there are 
abundant exercises with hints to their solution. 



Partial Differential Equations 

E.T.COPSON 

'Copson aims at both mathematicians and users of partial differential equations. 
There is, again, nothing available like this amalgam of pure theory and methods. 
It suddenly becomes clear as one reads how the various tricks fall into place as 
part of a surprisingly elegant whole ... The book will form a perfect advanced 
undergraduate or early postgraduate course in mathematics or physics . . . There 
is a good selection of examples in each chapter.' 
The Times Higher Education Supplement 



Cambridge University Press 



521 21480 7